252solngr2-01 2/7/01 Name: Class days and time: Please include this on what you hand in! Graded Assignment 2 2. A rental firm believes that that the average time that a backhoe is rented is 4.2 days. In order to verify this statement, a sample of rental records is taken, with the following results. 4, 2, 4, 3, 6, 2, 3, 2, 5, 3, 2, 4, 2, 2 You presumably now know that n 14 , x 3.1429 , s x2 1.6703 or s x 1.2924 . Test the hypothesis that the mean is 4.2 as follows: a) Use a test ratio and a 99% confidence level. State your null and alternative hypotheses. b) Use the test ratio to find an approximate p-value for the null hypothesis. Would you reject the null hypothesis if .01 ? .05 ? .10 ? c) Find a critical value or values (for the sample mean) for the null hypothesis and test it at the 99% confidence level. 3. (Ken Black) It is generally believed that 79% of US companies offer flexible scheduling . A survey of 415 accounting firms finds that 303 have flexible scheduling. Using a test ratio, test the hypothesis that the proportion of accounting firms that offer flexible hours is below that for other US firms. State your null and alternative hypotheses and find a p-value for the null hypothesis. If the significance level is 1% would you reject the null hypothesis? 4. Using a significance level of 1% and a critical value for the observed proportion, repeat 3. Solution: 2) You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion. a) Test Ratio: n 14 , x 3.1429 , s x2 1.6703 or s x 1.2924 , .01 . H 0 : 4.2 , H1 : 4.2 , so .005 4.2 . t n1 t 13 3.012 0 2 2 .005 s x2 x 0 3.1429 4.2 1.6703 0.3454 t 3.061 . Make a diagram showing an almost n 14 sx 0.3454 Normal curve with a mean at 0 and shaded 'reject' zones below -3.012 and above 3.012. Since t 3.061 is below -3.012, we reject H 0 . sx b) To find the p-value, we need (because this is a 2-sided test) pval 2Px 3.1429 2Pt 3.061 . Make a diagram showing an almost Normal curve with a mean at 0 and the p-value represented by a shaded area below -3.061 and another shaded area above 3.061. Use the 'df = 13' line of the t table. The 13 13 3.012 and t .001 3.852 . In other words numbers on the t table that are closest to 3.061 are t .005 Pt 3.012 .005 and Pt 3.852 .001 . Since 3.061 is between -3.012 and -3.852, we can say that .001 Pt 3.061 .005 . If we double this we have .002 pval .01 . So if .01, .05, or .10 we would reject the null hypothesis because the p-value is below all of these significance levels. c) Critical value: We need critical values for x above and below 4.2, so use x cv 0 t n1 s x 2 4.2 3.012 0.3454 4.2 1.040 or 3.160 to 5.24. Make a diagram showing an almost Normal curve with a mean at 4.2 and shaded 'reject' zones below 3.16 and above 5.24. Since x 3.1429 is below 3.16, we reject H 0 . 3) If we interpret the problem as saying that the proportion of firms that offer flexible hours is below 79%, we have stated an alternative hypothesis because it does not contain an equality. H 0 : p .79 H 1 : p .79 The problem also says that x 303 and n 415 , so that p x 303 .7301 and since p 0 .79, n 415 p0 q0 .79.21 0.01999 . Since the alternative hypothesis is H 1 : p .79 , we are worried n 415 about the proportion being too low, so pval P p .7301 . Under the null hypothesis p has the normal p p0 distribution with a mean of p 0 .79 and a standard deviation of p 0.01999 . So z , and p p .7301 .79 pval P p .7301 P z Pz 3.00 .5 .4987 .0013 . The rule on p-value(from .01999 the website) says if the p-value is less than the significance level (alpha) reject the null hypothesis; if the p-value is greater than or equal to the significance level, do not reject the null hypothesis. If .01 , .0013 is lower and we must reject the null hypothesis. 4) Since the alternative hypothesis is H 1 : p .79 , we are worried about the proportion being too low, so we want a critical value for p that is below .79. The critical value formula, pcv p0 z p must be 2 modified to give us a number below .79, so pcv p0 z p .79 2.327.01999 .79 .0465 .7435. Make a diagram with the mean at .79 showing the rejection zone is the area below .7435. Since p .7301 is below this critical value, reject H 0 . 2