Statistics Review 10A Name: Part I. Multiple choice. Circle the letter

advertisement
Statistics Review 10A
Name:
Part I. Multiple choice. Circle the letter corresponding to the best answer choice.
1. Complete the following sentence with the two best words
or phrases. As the sample size _____________, the
sampling distribution of the sample mean gets closer and
closer to ______________.
(a) increases;  .
(b) increases; a normal distribution.
(c) increases; zero.
(d) decreases; a skewed distribution.
(e) decreases; a uniform distribution.
standard deviation =
𝑥̅ −𝜇
𝑠
√𝑛
t = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
CI = 𝑥̅ ± 𝑡 ∗ (𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛)
2. Suppose a hypothesis test concerning a population mean was conducted with
H 0 :   25 and H a :   25 . A simple random sample of size n  32 was obtained and
the sample mean was x  23.2 and the standard deviation was s  7.3 . The standard
score in this hypothesis test is
(a) t = -17.98
(b) t = -7.89
(c) t = -1.39
(d) t = -0.25
(e) t = 1.39
3. Suppose a simple random sample of size n  42 is selected from a population with
mean   10 and standard deviation s  2.5 . If the sample mean is x  9.3 , the
standard score is
(a) t = -1.81
(b) t = -0.73
(c) t = -0.28
(d) t = -0.11
(e) t = -0.02
4. The margin of error for a 95% confidence interval is 2.8. If we decrease the confidence
level to 90%, the margin of error will be
(a) smaller than 2.8, making the interval wider
(b) larger than 2.8, making the interval narrower
(c) smaller than 2.8, making the interval narrower
(d) larger than 2.8, making the interval wider
5. In a test of H0:  = 4 against Ha:   4, a sample of size 100 produces x  4.5 and s = 2.5.
Thus the P-value (or observed level of significance) of the test is approximately equal to:
(a) 0.841
(b) 0.421
(c) 0.100
(d) 0.048
(e) 0.024
6. In a large population of adults, the mean IQ is 112 with a standard deviation of 20.
Suppose 200 adults are randomly selected for a market research campaign. The
distribution of the sample mean IQ is
(a) Exactly normal, mean 112, standard deviation 20.
(b) Approximately normal, mean 112, standard deviation 0.1.
(c) Approximately normal, mean 112, standard deviation 1.414.
(d) Approximately normal, mean 112, standard deviation 20.
7. The value of t* required for a 70% confidence interval with df = 14 is
(a) -0.5244
(b) 1.076
(c) 0.15
(d) 0.30
8. You want to compute a 90% confidence interval for the mean of a population with
unknown population standard deviation. The sample size is 30. The value of t* you
would use for this interval is
(a) 1.96
(b) 1.645
(c) 1.699
(d) .90
(e) 1.311
(f) None of the above
9. Suppose a hypothesis test is conducted concerning a population mean with
H 0 :   4.21 and H a :   4.21 . What does a P-value of 0.001 mean?
(a) There is no evidence to suggest the population mean is different from 4.21.
(b) There is strong evidence to suggest the population mean is different from 4.21.
(c) There is strong evidence to suggest the population mean is greater than 4.21.
(d) There is strong evidence to suggest the population mean is less than 4.21.
(e) There is strong evidence to suggest the individual population is not normal.
10. A simple random sample of size n  30 is obtained with a mean of 65 and a standard
deviation of 6. The 95% confidence interval for µ is:
(a) (53.0, 77.0)
(b) (62.8, 67.2)
(c) (61.7, 68.3)
(d) (59.3, 70.7)
(e) cannot be determined
11. An NCAA official claims that the average 10K time for students trying out for
college cross-country teams is 33 minutes. A Big 12 coach believes the true figure is
lower among Big 12 universities. He picks a simple random sample of 40 recruits and
calculates their mean 10K time is 32 minutes and 40 seconds, with a standard deviation
of 42 seconds. What is the appropriate test statistic and p-value? [HINT: 60 seconds = 1
minute]
a) z = -3.01, p = .0013
b) t = -3.01, p = .0045
c) z = -1.43, p = .0764
d) t = -3.01, p = .0022
e) There is insufficient information given to draw a conclusion.
Part II: Short Answer.
12. The manager of a pet supply store believes the reason many freshwater fish die in
home aquariums is because the pH level is greater than the acceptable level of 7.5. A
simple random sample of 48 home aquariums was obtained and the pH of each was
carefully measured. The sample mean was x  7.6 and the sample standard deviation
was s  0.35 . Use this information to determine if the mean pH of home aquariums is
greater than 7.5.
(a) State the null and the alternative hypothesis.
(b) Describe the sampling distribution.
(c) Find the test statistic and P-value.
(d) State your conclusion using an α = 0.05.
13. The level of dissolved oxygen in a river is an important indicator of the water’s
ability to support aquatic life. You collect water samples at 15 randomly chosen
locations along a stream and measure the dissolved oxygen. Here are your results in
milligrams per liter:
4.53, 5.04, 3.29, 5.23, 4.13, 5.50, 4.83, 4.40, 5.42, 6.38, 4.01, 4.66, 2.87, 5.73, 5.55
(a) Construct and interpret a 95% confidence interval for the mean dissolved oxygen
level for this stream.
(c) Does the interval support or reject a claim that the average amount of dissolved
oxygen for this stream is 4.0 mg/L?
Download