1. The mean of the population sampling distribution of x is given by
A) µ
B) σx
C) µx
D) x
2. A random sample of 90 observations produced a mean x = 25.9 and a standard deviation of s
= 2.7. Find a 95% confidence interval for the population mean µ.
A) (25.34, 26.46)
B) (25.34, 26.63)
C) (25.841, 25.958)
D) (25.63, 26.17)
3. What happens to the width of a confidence interval as the value of the confidence coefficient
is increased while the sample size is held fixed?
A) Increases
B) Decreases
C) Remains unchanged
4. Suppose you have selected a random sample of n = 7 measurements from a normal
distribution. Compare the standard normal z-values with the corresponding t-values if you were
forming a 95% confidence interval?
A) 1.943
B) 2.447
C) 1.895
D) 2.365
5. The __________ represents the status quo to the party performing the sampling experiment.
A) null hypothesis
B) alternative hypothesis
C) research hypothesis
D) test statistic
Be sure to use the Microsoft Equation Editor to show answers and calculations.
1) A university of Florida economist conducted a study of Virginia elementary school lunch
menus. During the stat-mandated testing period, school lunches averaged 863 calories.
(National Bureau of Economic Research, November 2002) The economist claimed that after
the testing period end, the average caloric content of Virginia school lunches dropped
significantly. Set up the null and alternative hypothesis to test the economist’s claim.
𝑁𝑢𝑙𝑙 𝐻𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝐻0 : 𝜇 => 863
𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 𝐻𝑦𝑝𝑜𝑡ℎ𝑠𝑖𝑠 𝐻𝑎 : 𝜇 < 863
2) A random sample of 64 observations produced the following summary statistics; 𝑥̅ = .323
and 𝑠 2 = .034. Test the null hypothesis that 𝜇 = .36 against the alternative hypothesis
that 𝜇 < .36, using 𝛼 = .10.
𝑧=
𝑧=
(𝑥̅ − 𝜇)
𝑠
√𝑛
(.323 − .36)
. 184
√64
𝑧=
−.037
. 023
𝑧 = −1.609
𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑧 𝑎𝑡 𝛼 = .10, 𝑜𝑛𝑒 𝑡𝑎𝑖𝑙𝑒𝑑 = −1.28
Because the observed z value (-1.609) is less than the critical z value (-1.28), we reject the null
hypothesis and accept the alternative hypothesis that there is a decrease in the average caloric
content.
3) Suppose you are interested in conducting the statistical test of 𝐻0 : 𝜇 = 200 against 𝐻𝑎 ∶ 𝜇 >
200, and you have decided to use the following decision rule: Reject 𝐻0 if the sample mean
of a random sample of 100 items is more than 215. Assume that the standard deviation of the
population is 80.
a) Express the decision rule in terms of 𝑧.
𝑧=
𝑧=
(𝑥̅ − 𝜇)
𝜎
√𝑛
(215 − 200)
80
√100
𝑧=
15
8
𝑧 = 1.875
Reject 𝐻0 if the observed z is more than 1.875
b) Find 𝛼, the probability of making a
𝛼 𝑓𝑜𝑟 𝑧(1.875) = .0304
Type I error, by using this decision
rule.
4) In order to compare the means of two populations, independent random samples of 400
observations are elected from each population, with the following results. Use a 95%
confidence interval to estimate the difference between the population means (𝜇1 − 𝜇2 ).
Interpret the confidence interval.
Sample 1
Sample 2
𝑥̅1 = 5,275 𝑥̅2 = 5,240
𝑠1 = 150
𝑠𝑥̅ 1 −𝑥̅2
𝑠2 = 200
𝑠12 𝑠22
=√ +
𝑛1 𝑛2
𝑠𝑥̅1 −𝑥̅2 = √
1502 2002
+
400
400
𝑠𝑥̅ 1 −𝑥̅ 2 = √156.25
𝑠𝑥̅ 1 −𝑥̅ 2 = 12.5
95% 𝐶𝐼 = (𝑥̅1 − 𝑥̅2 ) ± 𝐶𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑧 𝑋 𝑠𝑥̅1 −𝑥̅2
95% 𝐶𝐼 = 35 ± 1.96 𝑋 12.5
95% 𝐶𝐼 = 35 ± 24.5
95% 𝐶𝐼 = [10.5,59.5]
I am 95% confident that the true difference between the two means lies between 10.5 and 59.5.
5) Assume that 𝜎12 = 𝜎22 = 𝜎 2 . Calculate the pooled estimator of 𝜎 2 if 𝑠12 = 200, 𝑠22 =
180, 𝑛1 = 𝑛2 = 25
𝑆 2 𝑝𝑜𝑜𝑙𝑒𝑑 =
𝑆 2 𝑝𝑜𝑜𝑙𝑒𝑑 =
(𝑛1 − 1)𝑠12 + (𝑛2 − 1)𝑠22
𝑛1 + 𝑛2 − 2
(25 − 1)200 + (25 − 1)180
25 + 25 − 2
𝑆 2 𝑝𝑜𝑜𝑙𝑒𝑑 =
4800 + 4320
48
𝑆 2 𝑝𝑜𝑜𝑙𝑒𝑑 =
4800 + 4320
48
𝑆 2 𝑝𝑜𝑜𝑙𝑒𝑑 = 190