chapter 9 ex 3rd edition 9-2 a) The type I error probability is the probability of rejecting the null hypothesis (H0: μ = 12) when it is actually true. To find the type I error probability, we need to find the critical value that defines the critical region. In this case, the critical region is defined as x̄ < 11.5 kilograms. Since the sample size is small (n = 4), we can use the t-distribution instead of the normal distribution to find the critical value. The critical value can be found using a t-distribution table with degrees of freedom (df) equal to n-1 = 3 and a significance level of α. Using the t-distribution table, we can find the critical t-value that corresponds to a onetailed test with a significance level of α and df = 3. Let's assume α is set at 0.05. The critical t-value would be -2.365. This means that if the sample mean is less than -2.365 standard deviations below the null hypothesis mean (12), we reject H0. The type I error probability is the probability of observing a sample mean less than 11.5 kilograms when the true mean is equal to 12 kilograms. We can use the t-distribution formula to calculate the type I error probability: P(x̄ < 11.5 | μ = 12) = P(t < -2.365 | μ = 12) (b) The β (beta) is the probability of failing to reject the null hypothesis (H0) when the alternative hypothesis (H1) is true. To find β, we need to find the probability of observing a sample mean greater than or equal to 11.5 kilograms when the true mean is 11.25 kilograms. P(x̄ ≥ 11.5 | μ = 11.25) = P(t ≥ -2.365 | μ = 11.25) We can use the t-distribution formula to calculate β: β = P(t ≥ -2.365 | μ = 11.25) This value can be found using a t-distribution table with df = 3. 9-3 a) The type I error probability is the probability of rejecting the null hypothesis (H0: μ = 12) when it is actually true. To find the type I error probability, we need to find the critical value that defines the critical region. In this case, the critical region is defined as x̄ < 11.5 kilograms. Since the sample size is larger (n = 16), we can use the normal distribution instead of the t-distribution to find the critical value. The critical value can be found using a standard normal table with a significance level of α. Let's assume α is set at 0.05. The critical z-value would be -1.833. This means that if the sample mean is less than -1.833 standard deviations below the null hypothesis mean (12), we reject H0. The type I error probability is the probability of observing a sample mean less than 11.5 kilograms when the true mean is equal to 12 kilograms. We can use the standard normal distribution formula to calculate the type I error probability: P(x̄ < 11.5 | μ = 12) = P(z < -1.833 | μ = 12) (b) The β (beta) is the probability of failing to reject the null hypothesis (H0) when the alternative hypothesis (H1) is true. To find β, we need to find the probability of observing a sample mean greater than or equal to 11.5 kilograms when the true mean is 11.25 kilograms. P(x̄ ≥ 11.5 | μ = 11.25) = P(z ≥ -1.833 | μ = 11.25) We can use the standard normal distribution formula to calculate β: β = P(z ≥ -1.833 | μ = 11.25) This value can be found using a standard normal table. 9-4 In Exercise 9-2, the hypothesis test is to test H0: μ = 12 against H1: μ < 12 using a sample size of n = 4. The type I error probability, alpha, is specified to be 0.01. To find the boundary of the critical region, we need to find the critical value that corresponds to alpha = 0.01. In this case, we are using the t-distribution since the sample size is small (n = 4). The critical value can be found using a t-distribution table with 3 degrees of freedom (df = n - 1) and a significance level of α = 0.01. The critical t-value would be -3.182. This means that if the sample mean is less than -3.182 standard deviations below the null hypothesis mean (12), we reject H0. The critical region can be defined as x̄ < 11.49 kilograms (12 + (-3.182) * (0.5 / sqrt(4))). The boundary of the critical region is 11.49 kilograms. (Hence, the boundary of the critical region for the test with type I error probability of 0.01 and a sample size of 4 is x-critical = 12.5825. Any sample mean less than 12.5825 would result in rejection of the null hypothesis.) 9-5 In Exercise 9-2, the hypothesis test is to test H0: μ = 12 against H1: μ < 12 using a sample size of n = 4. The type I error probability, alpha, is specified to be 0.05. To find the boundary of the critical region, we need to find the critical value that corresponds to alpha = 0.05. In this case, we are using the t-distribution since the sample size is small (n = 4). The critical value can be found using a t-distribution table with 3 degrees of freedom (df = n - 1) and a significance level of α = 0.05. The critical t-value would be -1.833. This means that if the sample mean is less than -1.833 standard deviations below the null hypothesis mean (12), we reject H0. The critical region can be defined as x̄ < 11.93 kilograms (12 + (-1.833) * (0.5 / sqrt(4))). The boundary of the critical region is 11.93 kilograms. 9-5 To find the boundary of the critical region when the type I error probability (alpha) is 0.05, we need to find the critical value for a one-tailed, lower-tailed test with a significance level of 0.05. This can be done using a t-distribution table with 3 degrees of freedom (sample size - 1). Since the hypothesis is H0: mu = 12 against H1: mu < 12, we are performing a lower-tailed test. So, we need to find the critical value such that the area under the t-distribution to the left of that value is 0.05. From the t-distribution table, the critical value for a one-tailed, lower-tailed test with a significance level of 0.05 and 3 degrees of freedom is -2.492. Therefore, the boundary of the critical region is given by xbar < -2.492 * s/sqrt(n). Using the sample information from Exercise 9-2 (n = 4, s = 0.5, and xbar = the sample mean), we can calculate the critical region boundary as follows: xbar < -2.492 0.5 / sqrt(4) = -2.492 0.5 / 2 = -0.622 So, if the sample mean (xbar) is less than -0.622, we will reject the null hypothesis H0: mu = 12 and conclude that the mean elongation is less than 12 kilograms. เช็คตัวเลขใหม่ทุกอัน 9-6 a) The type I error probability α is the probability of rejecting the null hypothesis when it is actually true. To find α, we need to find the probability that the sample mean falls outside of the acceptance region when the true population mean is 100. Since the population is approximately normally distributed with mean 100 and standard deviation 2, the standard error of the mean is 2/sqrt(9) = 0.447. Using a z-score, we can find the proportion of the normal distribution outside of the acceptance region: z = (98.5 - 100) / 0.447 = -2.206 z = (101.5 - 100) / 0.447 = 2.206 Using a standard normal table, we find that the area to the left of -2.206 is 0.0132, and the area to the right of 2.206 is 0.0132. Therefore, α = 0.0132 + 0.0132 = 0.0264. b) To find β for the case where the true mean heat evolved is 103, we need to find the probability that the sample mean falls in the acceptance region when the true population mean is 103. Using the same standard error of the mean (0.447), we can find the z-score: z = (103 - 100) / 0.447 = 6.729 Using a standard normal table, we find that the area to the left of 6.729 is 1, which means that the sample mean will always be in the acceptance region when the true population mean is 103, so β = 0. c) To find β for the case where the true mean heat evolved is 105, we need to find the probability that the sample mean falls in the acceptance region when the true population mean is 105. Using the same standard error of the mean (0.447), we can find the z-score: z = (105 - 100) / 0.447 = 11.831 Using a standard normal table, we find that the area to the left of 11.831 is 1, which means that the sample mean will always be in the acceptance region when the true population mean is 105, so β = 0. This value of β is smaller than the one found in part (b) above because the difference between the true population mean (105) and the hypothesized mean (100) is larger than the difference between the true population mean (103) and the hypothesized mean (100). When the difference between the true population mean and the hypothesized mean is larger, the sample mean is less likely to fall in the acceptance region, so β is smaller. 9-7 For the given hypothesis test, H0: μ = 100 versus H1: μ ≠ 100, with a sample size of n=5 and an acceptance region of 98.5 ≤ x̄ ≤ 101.5, the type I error probability α can be calculated using the Z-score and Z-table. To calculate Z-score, we use the formula: Z = (x̄ - μ) / (σ / √n) where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. For the null hypothesis, μ = 100 and Z = (x̄ - 100) / (2 / √5) The Z-score corresponding to the lower limit of the acceptance region, 98.5, is calculated as follows: Z = (98.5 - 100) / (2 / √5) = -1.32 The Z-score corresponding to the upper limit of the acceptance region, 101.5, is calculated as follows: Z = (101.5 - 100) / (2 / √5) = 1.32 Using a Z-table, we can find the area under the standard normal distribution curve that corresponds to each Z-score. The area to the left of Z = -1.32 is 0.0921 and the area to the left of Z = 1.32 is 0.9053. Therefore, the type I error probability α is equal to 1 (0.9053 - 0.0921) = 0.0026. (b) To find the value of β for the case where the true mean heat evolved is 103, we can use the Z-score formula again. Z = (x̄ - 103) / (2 / √5) = (x̄ - 103) / 0.8944 The value of x̄ that corresponds to β = 0.05 is calculated using the Z-table. For a Z-score of 1.65, the area to the left is 0.95. Therefore, using the Z-score formula, we can calculate x̄ as follows: 103 + (1.65 * 0.8944) = 104.48 This means that if the true mean is 103, there is a 5% chance of obtaining a sample mean less than or equal to 104.48, resulting in a type II error. (c) To find the value of β for the case where the true mean heat evolved is 105, we can use the Z-score formula again. Z = (x̄ - 105) / (2 / √5) = (x̄ - 105) / 0.8944 The value of x̄ that corresponds to β = 0.05 is calculated using the Z-table. For a Z-score of 2.58, the area to the left is 0.99. Therefore, using the Z-score formula, we can calculate x̄ as follows: 105 + (2.58 * 0.8944) = 107.23 This means that if the true mean is 105, there is a 1% chance of obtaining a sample mean less than or equal to 107.23, resulting in a type II error. The value of β for the case where the true mean is 105 is smaller than the one found in part (b) above because the mean is farther away from the null hypothesis and the sample size is smaller. The smaller sample size makes it more difficult to detect a difference, resulting in a larger 9-8 (a) If the critical region is xbar = 185, then alpha is the probability of obtaining a sample mean of 185 or higher, given that the true mean foam height is 175. Using the standard normal table, we find that the z-score for xbar = 185 is (185 - 175) / (20/sqrt(10)) = 1.5. The probability of observing a z-score of 1.5 or greater is 0.0668, so the type I error probability alpha is 0.0668. (b) If the true mean foam height is 195 millimeters, then the type II error probability beta is the probability of not rejecting the null hypothesis H0: mu = 175 millimeters, given that the true mean is 195. The z-score for xbar = 195 is (195 - 175) / (20/sqrt(10)) = 3. The probability of observing a z-score less than 3 is 0.0013, so the type II error probability beta is 1 - 0.0013 = 0.9987. 9-15 For the hypothesis test in Exercise 9-8, H0: mu = 175 millimeters, H1: mu > 175 millimeters, with n = 10 and a critical value of xbar = 185 millimeters, the probability of accepting H0 (Type II error) can be calculated for various values of the true mean, mu. The operating characteristic (OC) curve can be constructed by plotting the probability of accepting H0 (1 - alpha) on the y-axis and the true mean, mu, on the x-axis for the given values of mu. Steps to calculate probability of accepting H0 for each value of mu: 1. Find the standard deviation of the sample mean, sigma/sqrt(n), where sigma = 20 millimeters (standard deviation of foam height) and n = 10. 2. For each value of mu, find the corresponding z-score using the formula z = (xbar mu)/(sigma/sqrt(n)). 3. Using the z-score, find the probability of accepting H0 using a standard normal table. For each value of mu, repeat these steps and plot the resulting probabilities on the y-axis versus the corresponding mu value on the x-axis to construct the OC curve. 9-9 (a) The conclusion would be that the sample data provide sufficient evidence to reject the null hypothesis H0: mu= 175 millimeters, in favor of the alternative hypothesis H1: mu >175 millimeters, since xbar = 190 millimeters falls in the critical region defined as xbar >=185 millimeters. (b) The probability of observing a sample average as large as 190 millimeters (or larger), if the true mean foam height was really 175 millimeters, can be found using the standard normal cumulative distribution function. The z-score for 190 millimeters is (190175)/(20/sqrt(10)) = 2.37. Therefore, the probability is given by the cumulative probability for z >= 2.37, which is 0.00889. This means that it is very UNUSUAL that we would observe a sample average as large as 190 millimeters when the true mean was really 175 millimeters. เช็คตัวเลขดีๆนะ มันประมาณหยาบไป 9-10. Repeat Exercise 9-8 assuming that the sample size is n =16 and the boundary of the critical region is the same. A consumer products company is formulating a new shampoo and is interested in foam height (in millimeters). Foam height is approximately normally distributed and has a standard deviation of 20 millimeters. The company wishes to test H0: mu= 175 millimeters versus H1: mu >175 millimeters, using the results of n = 16 samples. (a) Find the type I error probability alpha if the critical region is xbar =185. We need to find the critical value for x̄ at the alpha=0.05 level of significance, where x̄ is the sample mean. Since the critical region is xbar =185, the critical value is 185. To find the type I error probability, we need to find the probability of observing a sample mean of x̄ =185 or higher under the assumption that the true mean is mu= 175 millimeters (H0 is true). Using the z-score formula, we can find the z-score corresponding to the critical value 185: z = (185 - 175) / (20/√16) = (185 - 175) / (20/4) = 2.5 Using a standard normal table, the probability of observing a z-score of 2.5 or higher is 0.0062, so the type I error probability alpha is 0.0062. (b) What is the probability of type II error if the true mean is 195 millimeters? We need to find the probability of failing to reject H0 (i.e., accepting H0 as true) when the true mean is 195 millimeters. To do this, we find the z-score corresponding to the value of x̄ that is just below the critical value, and then find the corresponding probability using a standard normal table. z = (185 - 195) / (20/√16) = (185 - 195) / (20/4) = -2.5 The probability of observing a z-score of -2.5 or lower is 0.0062, so the probability of type II error (beta) if the true mean is 195 millimeters is 0.0062. 9-11. Consider Exercise 9-8, and suppose that the sample size is increased to n =16. a) Where would the boundary of the critical region be placed if the type I error probability were to remain equal to the value that it took on when n =10? The type I error probability alpha remains the same as before, so we can use the formula for the critical value for a one-tailed test with alpha = 0.05, a standard normal distribution, and a sample size of n =16: z = z(1-alpha) = z(0.95) = 1.833 This means that xbar must be greater than xbar_critical = 175 + 1.833 * (20/sqrt(16)) = 185.75 to reject the null hypothesis. b) Using n =16 and the new critical region found in part (a), find the type II error probability beta if the true mean foam height is 195 millimeters. The formula for the type II error probability beta is: beta = P(accept H0 | H1 is true) = P(xbar < xbar_critical | mu = 195) This can be calculated using a standard normal distribution table: beta = P(Z < (185.75 - 195)/(20/sqrt(16))) = P(Z < -2.86) = 0.002 (c) Compare the value of beta obtained in part (b) with the value from Exercise 9-8 (b). What conclusions can you draw? The value of beta obtained in part (b) is smaller than the value from Exercise 9-8 (b), meaning that the increased sample size has led to a smaller type II error probability. This is expected, as increasing the sample size generally results in a more powerful test and a lower probability of making a type II error. 9-12 a) To find alpha, we need to find the probability of observing a sample mean in the acceptance region when the true mean is not equal to 5 Volts. Since the standard deviation is 0.25 Volts and the sample size is 8, the standard error of the sample mean is 0.25/sqrt(8) = 0.10 Volts. Using a z-test, we can standardize the difference between the sample mean and the true mean and find the probability that the standardized difference falls within the critical region. The critical region is defined by 4.85 <= xbar <= 5.15, so the critical z-values are -1.8 and 1.8 respectively. Using a standard normal table, the area to the left of -1.8 is 0.0366 and the area to the right of 1.8 is 0.0366. The area between -1.8 and 1.8 is 1 - 2 * 0.0366 = 0.9268. So, alpha is 1 - 0.9268 = 0.0732, the probability of observing a sample mean in the acceptance region when the true mean is not equal to 5 Volts. (b) To find the power of the test, we need to find the probability of observing a sample mean outside of the acceptance region when the true mean is 5.1 Volts. Using the same calculation method as in (a), the standardized difference between the sample mean and the true mean is (5.1 - 5) / 0.10 = 1. Using a standard normal table, the area to the left of 1 is 0.8413. So, the power of the test is 1 - 0.8413 = 0.1587, the probability of observing a sample mean outside of the acceptance region when the true mean is 5.1 Volts. 9-13 (a) With a sample size of n = 16, the standard deviation of the sample mean decreases, resulting in a narrower acceptance region. To maintain the same alpha (type I error rate), the boundaries of the acceptance region need to be adjusted accordingly. To find the new boundaries, we can use the Z-distribution table to determine the Z-score corresponding to alpha/2, and then use that Z-score to calculate the new critical values. Let's assume that alpha remains the same as in Exercise 9-12, then: Z = (5.15 - 5)/(0.25/sqrt(16)) = 2.56 Z = (4.85 - 5)/(0.25/sqrt(16)) = -2.56 The new boundaries would be 4.95 <= xbar <= 5.05, as calculated using Z-scores and the standard deviation of the sample mean. (b) To find the power of the test for detecting a true mean output voltage of 5.1 Volts, we need to find the probability of not rejecting the null hypothesis when the true mean is 5.1 Volts. Let's calculate the Z-score for a true mean of 5.1 Volts: Z = (xbar - 5.1)/(0.25/sqrt(16)) = 1.28 Using the Z-distribution table, we find that the area under the curve to the right of Z = 1.28 is 0.8972. This is the probability of not rejecting the null hypothesis when the true mean is 5.1 Volts. The power of the test (1 - beta) is 1 - 0.8972 = 0.1028, meaning that there is a 10.28% chance of detecting a true mean output voltage of 5.1 Volts if it exists. 9-14 To find the acceptance region, we need to know the critical value for a two-tailed test with alpha = 0.05 and n = 8. Assuming a normal distribution for the sample mean, we can use the standard normal distribution table to find the critical value. The critical value for a two-tailed test with alpha = 0.05 and n = 8 is approximately 2.306. This means that the acceptance region is located 2.306 standard deviations from the mean on either side. The standard deviation for the sample mean is equal to the standard deviation of the population divided by the square root of the sample size, or (0.25/√8) = 0.125. The acceptance region is then located between (5 - 2.306 0.125) = 4.92 and (5 + 2.306 0.125) = 5.08. Therefore, the acceptance region is 4.92 <= xbar <= 5.08. 9-15 For the hypothesis test in Exercise 9-8, H0: mu = 175 millimeters, H1: mu > 175 millimeters, with n = 10 and a critical value of xbar = 185 millimeters, the probability of accepting H0 (Type II error) can be calculated for various values of the true mean, mu. The operating characteristic (OC) curve can be constructed by plotting the probability of accepting H0 (1 - alpha) on the y-axis and the true mean, mu, on the x-axis for the given values of mu. Steps to calculate probability of accepting H0 for each value of mu: 1. Find the standard deviation of the sample mean, sigma/sqrt(n), where sigma = 20 millimeters (standard deviation of foam height) and n = 10. 2. For each value of mu, find the corresponding z-score using the formula z = (xbar mu)/(sigma/sqrt(n)). 3. Using the z-score, find the probability of accepting H0 using a standard normal table. For each value of mu, repeat these steps and plot the resulting probabilities on the y-axis versus the corresponding mu value on the x-axis to construct the OC curve. 9-16 The power function of a hypothesis test is the probability of rejecting the null hypothesis (H0) when the alternative hypothesis (H1) is true. In other words, it is the probability of correctly detecting a difference when there is one. The power function of the test in Exercise 9-8 can be calculated as follows: For each value of the true mean (mu) from 178 to 199, calculate the z-score using the formula: z = (xbar - mu) / (standard deviation / square root of sample size) where xbar is the sample mean, mu is the true mean, standard deviation is 20 millimeters, and sample size is 10. Next, use the z-score to find the corresponding p-value using a standard normal distribution table. The p-value represents the probability of observing a sample mean as extreme or more extreme than xbar, assuming H0 is true. Finally, the power function can be calculated as 1 - p-value. This represents the probability of rejecting H0 when the true mean is equal to the value of mu. The resulting values of the power function can be plotted against the values of the true mean (mu) to create a power function plot. The plot will show how the power of the test changes as the true mean deviates from the null hypothesis. 9-17 (a) The null hypothesis H0: p=0.6 and the alternative hypothesis H1: p>0.6, where p is the proportion of voters who favor the use of oxygenated fuels. If exactly 60% of the voters favor the use of these fuels, then the mean number of voters who favor the use of these fuels is 500_0.6 = 300, and the standard deviation is sqrt(500_0.6*0.4) = 12.04. Using the normal approximation to the binomial distribution, we can calculate the probability of observing more than 400 voters who favor the use of these fuels as: P(X > 400 | p=0.6) = 1 - P(X <= 400 | p=0.6) Where X is the number of voters who favor the use of these fuels, and P(X <= 400 | p=0.6) can be calculated using the cumulative distribution function (CDF) of the normal distribution: P(X <= 400 | p=0.6) = CDF(Z), where Z = (400 - 300) / 12.04 = 1.66 Therefore, P(X > 400 | p=0.6) = 1 - CDF(1.66) Using a standard normal table, the CDF(1.66) can be found, which gives us the type I error probability alpha: alpha = 1 - CDF(1.66) (b) The type II error probability beta is the probability of accepting the null hypothesis H0 when the true proportion of voters who favor the use of oxygenated fuels is 0.75. The mean number of voters who favor the use of these fuels under H1 is 500_0.75 = 375, and the standard deviation is sqrt(500_0.75*0.25) = 12.5. Using the normal approximation to the binomial, we can calculate the probability of observing less than or equal to 400 voters who favor the use of these fuels as: beta = P(X <= 400 | p=0.75) = CDF((400 - 375) / 12.5) The value of beta can be found using a standard normal table. 9-18 a) The type I error probability is the probability of rejecting the null hypothesis when it is true, i.e., the probability of concluding that p > 0.3 when p = 0.3. To calculate this, we use the cumulative distribution function (CDF) of the binomial distribution with parameters n = 10 and p = 0.3. The type I error probability is given by: P(X <= 1 | p = 0.3) = 1 - P(X > 1 | p = 0.3) = 1 - P(X >= 2 | p = 0.3) = 1 - CDF(2, 10, 0.3) = 1 - 0.2955 (b) The type II error probability is the probability of accepting the null hypothesis when it is false, i.e., the probability of not rejecting the null hypothesis when p = 0.2. This is given by: P(X <= 1 | p = 0.2) = CDF(1, 10, 0.2) = 0.125 (c) The power of the test is 1 - beta, where beta is the type II error probability. Hence, the power of the test is: power = 1 - beta = 1 - 0.125 = 0.875 9-19 a) The type I error probability for this procedure can be found by computing the probability of obtaining a sample that falls outside the acceptance region (4 to 8 graduates) when the true proportion is p = 0.4. Since the sample size is small (n = 15), we will use the normal approximation to the binomial distribution. Let p̂ be the sample proportion of graduates. Then, the standard deviation of p̂ is sqrt(p(1-p)/n), where p = 0.4. The critical region is defined by p̂ < 4/15 or p̂ > 8/15. To find the type I error probability, we need to find the area under the standard normal distribution curve to the left of z = (4/15 - 0.4)/(sqrt(0.4(1-0.4)/15)) and to the right of z = (8/15 - 0.4)/(sqrt(0.4(1-0.4)/15)). Using a standard normal table, we find that the area to the left of z = -1.47 is 0.0759, and the area to the right of z = 1.47 is 0.0759. So, the type I error probability is alpha = 2 * 0.0759 = 0.1518. (b) To find the type II error probability, we need to find the probability of obtaining a sample that falls within the acceptance region (4 to 8 graduates) when the true proportion is p = 0.2. The standard deviation of p̂ is still sqrt(p(1-p)/n), where p = 0.2. We need to find the area under the standard normal distribution curve between z = (4/15 - 0.2)/(sqrt(0.2(10.2)/15)) and z = (8/15 - 0.2)/(sqrt(0.2(1-0.2)/15)). Using a standard normal table, we find that the area between z = -0.69 and z = 1.08 is 0.8291. So, the type II error probability is beta = 1 - 0.8291 = 0.1709. 9-20 (a) To test the hypothesis, we can use a one-sample z-test with the null hypothesis H0: μ = 100 and the alternative hypothesis H1: μ > 100. The z-score for the sample mean of 98°F is calculated as follows: z = (98 - 100) / (2 / sqrt(9)) = -1.7888543819998317 Since the z-score is negative, it falls in the left tail of the standard normal distribution. We can use a standard normal table to find the area to the left of the z-score, which is the Pvalue. The P-value is equal to 0.0404. Since the P-value is less than the significance level α = 0.05, we reject the null hypothesis and conclude that the water temperature is not acceptable. (b) The P-value for the test is 0.0404. (c) To find the probability of accepting the null hypothesis at α = 0.05 when the true mean temperature is 104°F, we need to calculate the z-score for the sample mean in this scenario. z = (104 - 100) / (2 / sqrt(9)) = 1.7888543819998317 The area to the right of the z-score is 0.0404, which is the probability of accepting the null hypothesis (H0: μ = 100) at α = 0.05 when the true mean temperature is 104°F. revise 8-9 3rd editio for 9-21 To find the 95% two-sided confidence interval on the true mean yield, we will use the formula for a t-confidence interval for the mean: x̄ ± t* (s/√n) where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t is the t-critical value for the desired level of confidence and degrees of freedom (df = n1). For a 95% confidence interval, with df = 5-1 = 4, the t-critical value from a t-distribution table is 2.776. We can calculate the sample mean and standard deviation as follows: x̄ = (91.6 + 88.75 + 90.8 + 89.95 + 91.3) / 5 = 90.724 s = sqrt(((91.6 - x̄)^2 + (88.75 - x̄)^2 + (90.8 - x̄)^2 + (89.95 - x̄)^2 + (91.3 - x̄)^2) / (5-1)) = 0.7944 Finally, substituting the values in the formula, we have: x̄ ± t (s/√n) = 90.724 ± 2.776 (0.7944 / sqrt(5)) = 90.724 ± 2.776 * 0.4115 = (89.05, 92.4) So the 95% two-sided confidence interval for the true mean yield is (89.05, 92.4). This means we can be 95% confident that the true mean yield of the chemical process lies between 89.05% and 92.4%. 9-21 (a) To test the hypothesis that the mean yield is not 90%, we will perform a onesample t-test. The null hypothesis is H0: μ = 90% and the alternative hypothesis is H1: μ ≠ 90%. The test statistic t is calculated as follows: t = (x̄ - μ) / (s / √n) = (91.1 - 90) / (3 / √5) = 2.67 The degrees of freedom for this test are n - 1 = 4. Using a t-distribution table with 4 degrees of freedom and a two-tailed test with alpha = 0.05, the critical values are ±2.776. Since the calculated t-value of 2.67 is not within the critical region, we fail to reject the null hypothesis. (b) The P-value is the probability of obtaining a test statistic as extreme or more extreme than the one we calculated under the null hypothesis. The P-value can be calculated using a t-distribution table or using software. With a calculated t-value of 2.67 and 4 degrees of freedom, the P-value is approximately 0.06. (c) To detect a true mean yield of 85% with probability 0.95, we need to find the sample size required to achieve a desired power of 0.95. The formula for power is given by (1 - β), where β is the type II error probability. To solve for n, we need to find the t-value corresponding to a power of 0.95 with 4 degrees of freedom (since the sample size is unknown) and a two-tailed test with alpha = 0.05. Then, we can use the formula for tvalue given above to solve for n. (d) The type II error probability β is the probability of failing to reject the null hypothesis when it is false. To find β for a true mean yield of 92%, we need to find the t-value that corresponds to a type II error probability of β with 4 degrees of freedom and a two-tailed test with alpha = 0.05. Then, we can use the formula for t-value given above to see if the calculated t-value is within the critical region. If it is not within the critical region, we would fail to reject the null hypothesis, and the type II error probability would be β. (e) The decision made in part (c) is not directly comparable to the 95% CI on mean yield that was constructed in Exercise 8-7. The 95% CI gives an interval estimate of the population mean yield with a certain degree of confidence, while the hypothesis test in part (c) gives a decision about the population mean yield based on a single sample. However, if the 95% CI does not contain the value of 90% (the null hypothesis), this would suggest evidence against the null hypothesis, and we would reject the null hypothesis in a hypothesis test with alpha = 0.05. 9-22 a) To test the hypothesis H0: μ = 3 vs. H1: μ ≠ 3 with a significance level of α = 0.05, we need to find the critical value of the test statistic. Since we have the population standard deviation (σ = 0.9) and the sample size is large enough (n = 15), we can use the z-test statistic: z = (x̄ - μ) / (σ/√n) = (2.78 - 3) / (0.9/√15) = -2.13 Using a standard normal table, we can find the p-value for this test statistic: p-value = 2 P(Z < -2.13) = 2 0.0156 = 0.0312 Since the p-value is greater than the significance level (α = 0.05), we fail to reject the null hypothesis. We do not have enough evidence to suggest that the mean wear is different from 3. (b) The power of the test is defined as the probability of rejecting the null hypothesis when it is false. The power of the test when μ = 3.25 can be calculated as: z = (x̄ - μ) / (σ/√n) = (2.78 - 3.25) / (0.9/√15) = -3.33 p-value = 2 P(Z < -3.33) = 2 0.0005 = 0.001 Since the p-value is smaller than the significance level (α = 0.05), we reject the null hypothesis. The power of the test is the probability of making this correct decision and is equal to 1 - β, where β is the type II error probability. Therefore, the power of the test when μ = 3.25 is 1 - β = 1 - 0.0005 = 0.999. (c) To determine the required sample size to detect a true mean of 3.75 with a power of 0.9, we can use the formula for the sample size of a one-sided z-test: n = (z1 - z2)^2 * σ^2 / (μ1 - μ0)^2 where z1 and z2 are the critical values for the desired power and significance level, μ1 is the true mean, and μ0 is the null hypothesis mean. Solving for n, we have: n = (2.33 - 1.64)^2 * 0.9^2 / (3.75 - 3)^2 = 83.3 Therefore, a sample size of n = 84 is required to detect a true mean of 3.75 with a power of 0.9 at the α = 0.05 significance level. 9-23 a) To test H0: μ = 155 versus H0: μ ≠ 155 using α = 0.01, we can use a one-sided ztest. The test statistic is calculated as: z = (x̄ - μ) / (σ / √n) = (154.2 - 155) / (1.5 / √10) = -0.67 Since the test is one-tailed, the critical value of z at α = 0.01 is 2.33. Because the calculated z value of -0.67 is not greater than the critical value, we fail to reject the null hypothesis. b) The P-value for this test is the probability of observing a test statistic as extreme or more extreme than the one calculated, given that the null hypothesis is true. In this case, the P-value is calculated as the area under the standard normal curve to the left of -0.67. Using a standard normal table or calculator, the P-value can be calculated to be approximately 0.25. c) The β-error is the probability of failing to reject the null hypothesis when the alternative hypothesis is true. In this case, β is the probability of failing to reject H0: μ = 155 when μ = 150. The β-error is calculated using a z-test, with the test statistic: z = (x̄ - μ) / (σ / √n) = (154.2 - 150) / (1.5 / √10) = 1.6 Using a standard normal table or calculator, β can be calculated to be approximately 0.06. d) To find the value of n required to get β < 0.1 when μ = 150, we can use the equation for the test statistic and solve for n: n = (z * (σ / √n) / (μ - x̄))^2 Substituting the values for z = 1.96 (the critical value at α = 0.01 for a two-tailed test), σ = 1.5, μ = 150, and x̄ = 154.2, we get: n = (1.96 * (1.5 / √n))^2 / (150 - 154.2)^2 Solving for n, we get n = approximately 37. 9-24 a) To test the claim that battery life exceeds 40 hours, we can perform a one-tailed test with the null hypothesis H0: μ = 40 and alternative hypothesis H1: μ > 40, where μ is the true mean life of the battery. Using α = 0.05, the critical value for a one-tailed test with 9 degrees of freedom (df = n 1) can be found using a t-distribution table. The critical value is 1.833. We calculate the t-statistic as follows: t = (x̄ - μ) / (σ / √n) = (40.5 - 40) / (1.25 / √10) = 4 Since 4 > 1.833, we reject the null hypothesis H0 and conclude that there is evidence to support the claim that battery life exceeds 40 hours. (b) The P-value is the probability of observing a test statistic as extreme as the one we obtained, or more extreme, assuming the null hypothesis is true. The P-value can be calculated as: P-value = P(t > 4, df = 9) = 1 - P(t < 4, df = 9) Using a t-distribution table, the P-value is approximately 0.0002, which is much less than α = 0.05. (c) The β-error is the probability of failing to reject the null hypothesis when the alternative hypothesis is true. If the true mean life is 42 hours, then the β-error is calculated as follows: β = P(t < 1.833, df = 9 | μ = 42) = P(t > -1.833, df = 9 | μ = 42) Using a t-distribution table, the β-error is approximately 0.08. (d) To ensure that β does not exceed 0.10 when the true mean life is 44 hours, we need to calculate the minimum sample size required. Let's call the minimum sample size n. We can use the equation for t-statistic to solve for n: t = (x̄ - μ) / (σ / √n) = 1.833 Rearranging, we get: n = (tσ / √(x̄ - μ))^2 Substituting the known values: n = (1.833 * 1.25 / √(44 - 40.5))^2 = (2.291)^2 = 5.288 ≈ 6 Therefore, a sample size of n = 6 would be required to ensure that β does not exceed 0.10. (e) To answer the question in part (a) by calculating an appropriate confidence bound on life, we can use a t-confidence interval. A 95% t-confidence interval for the mean life of the battery is calculated as follows: CI = x̄ ± t (σ / √n) = 40.5 ± 1.833 (1.25 / √10) = (37.77, 43.23) Since the confidence interval does not contain 40, we conclude that there is evidence to support the claim that battery life exceeds 40 hours. 9-25 a) Testing the hypothesis: H0: μ = 3500 psi (the mean tensile strength is equal to 3500 psi) Ha: μ ≠ 3500 psi (the mean tensile strength is not equal to 3500 psi) Significance level: α = 0.01 We can use a two-sample t-test to test this hypothesis: t = (x̄ - μ) / (s / √n) Where x̄ is the sample mean (3250 psi), μ is the population mean (3500 psi), s is the sample standard deviation (not known), and n is the sample size (12). Since we don't know the population standard deviation σ, we need to estimate it using the sample standard deviation: s = sqrt[(sum(x_i - x̄)^2) / (n - 1)] where x_i are the observations. We can calculate the t-statistic: t = (3250 - 3500) / (s / sqrt(12)) We don't have the exact value for s, so we'll use the standard deviation estimate 60 psi. t = (3250 - 3500) / (60 / sqrt(12)) = -5.16 We can use a t-distribution table with 11 degrees of freedom (n - 1 = 12 - 1 = 11) to find the p-value: p-value = P(t < -5.16) = 0.000079 Since the p-value is less than the significance level (0.000079 < 0.01), we reject the null hypothesis. There is evidence to suggest that the mean tensile strength is not equal to 3500 psi. (b) Smallest level of significance at which you would be willing to reject the null hypothesis: To find the smallest level of significance at which you would be willing to reject the null hypothesis, we can find the critical value at which the p-value becomes greater than the significance level. For example, at α = 0.05, the critical value would be: t_critical = t_distribution_table(0.05, 11 degrees of freedom) = -1.833 Since the calculated t-statistic (-5.16) is less than the critical value (-1.833), we would still reject the null hypothesis even at a significance level of 0.05. (c) Confidence Interval: To answer the question in part (a) with a two-sided confidence interval, we can calculate a 95% confidence interval (α = 0.05) for the mean tensile strength: CI = x̄ ± t_critical * (s / sqrt(n)) Using the same standard deviation estimate of 60 psi and the same sample mean x̄ = 3250 psi, we get: CI = 3250 ± t_critical (60 / sqrt(12)) = 3250 ± 1.833 (60 / sqrt(12)) = (3157, 3343) The 95% confidence interval for the mean tensile strength is (3157, 3343) psi. Since 3500 psi is not within the confidence interval, we have evidence to suggest that the mean tensile strength is not equal to 3500 psi. 9-26 To solve this problem, we need to find the sample size required to detect a true mean strength of 3500 psi with a power of 0.8. Let's use the following notation: μ0: null hypothesis mean strength = 3500 psi μA: alternative hypothesis mean strength ≠ 3500 psi α: level of significance β: type II error probability n: sample size We know that: Power = 1 - β, where Power is the probability of correctly rejecting the null hypothesis if the alternative is true. Power = 0.8 (the desired value) α = 0.01 (from Exercise 9-25) σ = 60 psi (known standard deviation of tensile strength) We want to find the minimum sample size (n) that will give us a power of 0.8. To do this, we can use the formula for the sample size needed to achieve a certain level of power: n = (z1 - z2)^2 * σ^2 / (μA - μ0)^2 where z1 and z2 are the Z-scores corresponding to α/2 and β, respectively. To find z1 and z2, we can use a Z-table or use a statistical software package. Let's use a Z-table: For α/2 = 0.005, z1 = 2.58 For β = 0.2, z2 = 0.84 Substituting the values in the formula: n = (2.58 - 0.84)^2 * 60^2 / (μA - 3500)^2 Since μA is unknown, we can choose a value that makes the calculation easier. For example, let's choose μA = 3700 psi. Then: n = (2.58 - 0.84)^2 * 60^2 / (3700 - 3500)^2 = 49.44 Rounding up to the nearest integer, the sample size required to detect a true mean strength of 3500 psi with power 0.8 is n = 50. 9-27 a) The null hypothesis H0: μ = 100 and alternative hypothesis H1: μ ≠ 100 are tested using a one-sample z-test. The test statistic is calculated as follows: z = (x̄ - μ) / (σ/√n) = (102.2 - 100) / (4/√8) = 2.8 Since we have a two-tailed test, the critical value at α = 0.05 is ±1.96. Since z = 2.8 > 1.96, we reject the null hypothesis H0. (b) The power of the test can be calculated as 1 - β, where β is the probability of failing to reject the null hypothesis when it is false. If the true mean speed is μ = 95, the test statistic becomes: z = (102.2 - 95) / (4/√8) = 5.55 Since z > 1.96, the null hypothesis is rejected, and the power of the test is 1 - β = 1. (c) To find the sample size required to detect a true mean speed as low as 95 with a power of at least 0.85, we use the formula for the sample size n: n = (z1-α/2 + z1-β)^2 * σ^2 / (μ1 - μ0)^2 where z1-α/2 = 1.96, z1-β = norm.ppf(1-0.15), μ1 = 102.2, and μ0 = 95. n = (1.96 + norm.ppf(1-0.15))^2 * (4^2) / (102.2 - 95)^2 n = 44.7 The sample size required is approximately 45. (d) To answer the question in part (a) with a one-sided confidence bound, we can calculate a one-sided upper confidence bound on the mean speed. The confidence bound can be calculated as follows: x̄ + zα/2 σ/√n = 102.2 + 1.96 4/√8 = 106.82 The one-sided upper confidence bound on the mean speed is 106.82 meters per second, which is higher than 100 meters per second. This suggests that there is evidence to support the claim that the mean speed exceeds 100 meters per second. 9-28 a) To test the hypotheses H0: mu = 1.5 versus H1: mu ≠ 1.5 using alpha = 0.01, we can use a Z-test for the mean. The test statistic Z is calculated as Z = (x̄ - μ) / (σ / √n), where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this case, Z = (1.4975 - 1.5) / (0.01 / √25) = -2.5. Since Z < -2.576, which is the critical value for alpha = 0.01 with 24 degrees of freedom, we reject the null hypothesis H0. b) To compute the power of the test if the true mean diameter is 1.495 inches, we need to find the probability of rejecting the null hypothesis H0 if the true mean is 1.495 inches. The test statistic Z is Z = (1.4975 - 1.495) / (0.01 / √25) = 1. The power of the test is the complement of the probability of a type II error, which is the probability of failing to reject H0 when H0 is false. The power of the test can be calculated using a Z-distribution table or a software package. c) To find the sample size required to detect a true mean diameter as low as 1.495 inches if we wanted the power of the test to be at least 0.9, we can use the formula for sample size n = (Z1-α/2 + Z1-β)^2 σ^2 / (μ0 - μ1)^2, where Z1-α/2 and Z1-β are the critical values of the Zdistribution corresponding to alpha/2 and beta, respectively. Solving for n, we get n = (2.576 + 1.282)^2 0.01^2 / (1.5 - 1.495)^2 = (3.858)^2 * 0.0001 / 0.005^2 = approximately 65. d) To answer the question in part (a) by constructing a two-sided confidence interval on the mean diameter, we can use the formula for the confidence interval as x̄ ± Zα/2 (σ / √n), where x̄ is the sample mean, Zα/2 is the critical value of the Z-distribution corresponding to alpha/2, and n is the sample size. In this case, the 95% two-sided confidence interval on the mean diameter is 1.4975 ± 2.576 (0.01 / √25) = [1.4867, 1.5083]. Since the interval does not contain the value 1.5, we reject the null hypothesis H0 and conclude that the true mean diameter is different from 1.5. 9-29 a) To test the hypotheses H0: mu = 1.5 versus H1: mu ≠ 1.5 using alpha = 0.01, we can use a Z-test for the mean. The test statistic Z is calculated as Z = (x̄ - μ) / (σ / √n), where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. In this case, Z = (1.4975 - 1.5) / (0.01 / √25) = -2.5. Since Z < -2.576, which is the critical value for alpha = 0.01 with 24 degrees of freedom, we reject the null hypothesis H0. b) To compute the power of the test if the true mean diameter is 1.495 inches, we need to find the probability of rejecting the null hypothesis H0 if the true mean is 1.495 inches. The test statistic Z is Z = (1.4975 - 1.495) / (0.01 / √25) = 1. The power of the test is the complement of the probability of a type II error, which is the probability of failing to reject H0 when H0 is false. The power of the test can be calculated using a Z-distribution table or a software package. c) To find the sample size required to detect a true mean diameter as low as 1.495 inches if we wanted the power of the test to be at least 0.9, we can use the formula for sample size n = (Z1-α/2 + Z1-β)^2 σ^2 / (μ0 - μ1)^2, where Z1-α/2 and Z1-β are the critical values of the Zdistribution corresponding to alpha/2 and beta, respectively. Solving for n, we get n = (2.576 + 1.282)^2 0.01^2 / (1.5 - 1.495)^2 = (3.858)^2 * 0.0001 / 0.005^2 = approximately 65. d) To answer the question in part (a) by constructing a two-sided confidence interval on the mean diameter, we can use the formula for the confidence interval as x̄ ± Zα/2 (σ / √n), where x̄ is the sample mean, Zα/2 is the critical value of the Z-distribution corresponding to alpha/2, and n is the sample size. In this case, the 95% two-sided confidence interval on the mean diameter is 1.4975 ± 2.576 (0.01 / √25) = [1.4867, 1.5083]. Since the interval does not contain the value 1.5, we reject the null hypothesis H0 and conclude that the true mean diameter is different from 1.5. 9-30 CD