Download 1. An estimator is consistent if, as the sample size decreases, the

1. An estimator is consistent if, as the sample size decreases, the value of the estimator
approaches the value of the parameter estimated. __F__ T/F
Answer: False
Explanation: An estimator is consistent if, as the sample size increases, the value of the
estimator approaches the value of the parameter estimated.
2. For a specific confidence interval, the larger the sample size, the smaller the maximum
error of estimate will be. __T___ T/F
Answer: True
3. When we reject the null hypothesis, we are certain that the null hypothesis is false.
__F_ T/F
Answer: False
Explanation: When we reject the null hypothesis, we may conclude that the null
hypothesis is false at certain level of significance based on the sample information. But
there is a chance that the decision may be wrong.
4. The alternative hypothesis, sometimes referred to as the research hypothesis, is
supported by using the sample evidence to contradict the null hypothesis. _T_ T/F
Answer: True
5. α is the measure of the area under the curve of the standard score that lies in the
rejection region for the null hypothesis. _T_ T/F
Answer: True
6. Rejection of a null hypothesis that is false is a Type II error. _F_ T/F
Answer: False
Explanation: Acceptance of a null hypothesis that is false is a Type II error.
7. The maximum error of estimate is controlled by three factors: level of confidence,
sample size, and standard deviation. __T__ T/F
Answer: True
8. You are constructing a 95% confidence interval using the information: n = 50, x-bar=
54.3, s = 2.1 and E = 0.65. What is the value of the middle of the interval?
A. 2.1
B. 54.3
C. 50
D. 0.95
Answer: B. 54.3
Explanation: The value of the middle of the interval is x-bar = 54.3.
9. Which of the following would be the correct hypotheses for testing the claim that the
mean life of a battery for a cellular phone (while the phone is left on) is at least 24 hours?
A.
B.
C.
D.
H0: µ ≥ 24 and Ha: µ < 24
H0: µ = 24 and Ha: µ ≠ 24
H0: µ ≤ 24 vs. Ha: µ > 24
H0: µ > 24 vs. Ha: µ ≥ 24
Answer: C. H0: µ ≤ 24 vs. Ha: µ > 24
Explanation: The alternative hypothesis should be the mean life of a battery for a cellular
phone (µ) is at least 24 hours (µ > 24).
10. Which of the following would be the alternative hypothesis in testing the claim that
the mean distance students commute to campus is no less than 8.2 miles?
A. Ha: µ ≠ 8.2
B. Ha: µ > 8.2
C. Ha: µ < 8.2
D. Ha: µ ≤ 8.2
Answer: B. Ha: µ > 8.2
Explanation: The alternative hypothesis should be the mean distance students commute to
campus (µ) is no less than 8.2 miles (µ > 8.2).
Part II. Short Answers & Computational Questions
1. A study was conducted to estimate the mean amount spent on birthday gifts for a
typical family having two children. A sample of 175 was taken, and the mean amount
spent was $250. Assuming a standard deviation equal to $45, find the 95% confidence
interval for µ?, the mean for all such families. (4 pts)
If x-bar denotes the sample mean and σ denotes the standard deviation, the 95%
confidence interval for µ is (x-bar – 1.96 * SE, x-bar + 1.96 * SE), where SE = σ/√n
Here it is given that
x-bar = 250, σ = 45 and n = 175
Therefore, SE = σ/√n = 45/√175 = 3.4017
The 95% CI for the mean amount spent is given by
(x-bar – 1.96 * SE, x-bar + 1.96 * SE)
= (250 – 1.96 * 3.4017, 250 + 1.96 * 3.4017)
= (250 – 6.67, 250 + 6.67)
= ($243.33, $256.67)
2. What sample size would be needed to estimate the population mean to within one-third
standard deviation with 95% confidence? (4 pts)
If σ denotes the standard deviation, the margin of error is given as σ/3
Thus the sample size needed to estimate the population mean to within one-third standard
deviation with 95% confidence is
n = (z * σ /Margin of error)^2 = [1.96 * σ /( σ/3)]^2 = (1.96 * 3)^2 = 34.57
Hence a sample size of 35 is required.
3. A 95% confidence interval estimate for a population mean was computed to be (63.4 to
68.2). Determine the mean of the sample, which was used to determine the interval
estimate. (4 pts)
Mean = (63.4 + 68.2)/2 = 131.6/2 = 65.8
Thus the mean of the sample = 65.8
4. In testing the hypothesis, H0: µ ≥31.5 and Ha: µ < 31.5, using the p-value approach, a
p-value of 0.0409 was obtained. If σ = 8.7, find the sample mean which produced this pvalue given that the sample of size n = 50 was randomly selected. (4 pts)
The z- score corresponding to a p- value (Left-tailed test) of 0.0409 is -1.74
We have, the z score, z= (x-bar - µ)/SE, where SE = σ/√n
Therefore,
x-bar = µ + z * SE
= 31.5 + (-1.74)(8.7/√50)
= 31.5 – 2.14
= 29.36
5. To test the null hypothesis that the average lifetime for a particular brand of bulb is 725
hours versus the alternative that the average lifetime is different from 725 hours, a sample
of 100 bulbs is used. If the standard deviation is 50 hours and α is equal to 0.01, what
values for x-bar will result in rejection of the null hypothesis. (4 pts)
Here it is given that,
σ = 50, n = 100
Therefore,
SE = σ/√n = 50/√100 = 5
The critical value of z corresponding to α = 0.01 (Two-tailed) is 2.576
The test statistic for testing the null hypothesis is
z = (x-bar – µ)/SE
We reject the null hypothesis if z < -2.576 or z > 2.576
To reject the null hypothesis, we need (x-bar – 725)/5 < -2.576 or (x-bar – 725)/5 > 2.576
x-bar < 725 - 5 * 2.576 or x-bar > 725 + 5 * 2.576
x-bar < 725 – 12.88 or x-bar > 725 + 12.88
x-bar < 712.12 hours or x-bar > 737.88 hours
6. Describe the action that would result in a type I error and a type II error if each of the
following null hypotheses were tested. (4 pts)
a. H0: There is no waste in US Defense Department spending
b. H0: This fast-food menu is not low fat
(a) If there really is no waste in US Defense Department spending and if H0 is rejected
then a Type I error is committed. If there is waste in US Defense Department spending
and if we fail to reject H0 then a Type II error is committed.
(b) If the fast food menu is really not low fat and if H0 is rejected then a Type I error is
committed. If the fast food menu is low fat and if we fail to reject H0 then a Type II error
is committed.
7. By measuring the amount of time it takes a component of a product to move from one
workstation to the next, an engineer has estimated that the standard deviation is 5.1
seconds. (4 pts)
a. How many measurements should be made in order to be 95% certain that the
maximum error of estimation will not exceed 1.5 seconds?
b. What sample size is required for a maximum error of 2.5 seconds?
(a) Here it is given that, the margin of error = 1.5 and standard deviation, σ = 5.1
Thus the number of measurements should be made in order to be 95% certain that the
maximum error of estimation will not exceed 1.5 seconds is
n = (z * σ /Margin of error)^2 = (1.96 * 5.1/1.5)^2 = 44.41
Thus a sample size of 45 is required.
(a) Here it is given that, the margin of error = 2.5
Thus the sample size required for a maximum error of 2.5 seconds is
n = (z * σ /Margin of error)^2 = (1.96 * 5.1/2.5)^2 = 15.99
Thus a sample size of 16 is required.
8. Determine the critical region and critical values for z that would be used to test the null
hypothesis at the given level of significance, as described in each of the following: (6pts)
a. H0: µ ≤ 51 and Ha: µ > 51, α = 0.10
b. H0: µ ≥ 28 and Ha: µ < 28, α = 0.01
c. H0: µ = 93 and Ha: µ ≠ 93, α = 0.05
a) Since the alternative hypothesis is Ha: µ > 51, the critical region is the region in the
right tail (upper tail). At α = 0.10, the critical value is z = 1.28.
Thus the critical region is: Reject H0 if the calculated value of the test statistic > 1.28
b) Since the alternative hypothesis is Ha: µ < 28, the critical region is the region in the left
tail (lower tail). At α = 0.01, the critical value is z = -2.33.
Thus the critical region is: Reject H0 if the calculated value of the test statistic < -2.33
c) Since the alternative hypothesis is Ha: µ ≠ 93, the critical region is the region in the two
tails. At α = 0.05, the critical value is z = ± 1.96.
Thus the critical region is: Reject H0 if the calculated value of the test statistic < -1.96 or
if the calculated value of the test statistic > 1.96