Practice Problems for Section 7.1 (without using Minitab)
ANSWERS
1) A simple random sample of five female basketball players is selected. Their heights (in cm) are 170,
175, 169, 183, and 177. What is the standard error of the mean of these height measurements?
The sample st dev s = 5.6745, so
2) A sample of size n = 27 is used to conduct a significance test for H0: μ = 75 versus Ha: μ > 75. The test
statistic is t = 3.45.
What are the degrees of freedom for this test statistic?
df = n – 1 = 26
If we let α = 0.05, will we reject Ho and support Ha? Why or why not? (include a sketch)
The cut-off value of t for the rejection region is 1.706. If t ≥ 1.706, then, we will be able to reject Ho.
Our value of t was 3.45, and so we ARE in the rejection region. We have enough evidence to support
Ha. We conclude that the population mean is greater than 75.
3) A simple random sample of 20 third-grade children from a certain school district is selected, and each
is given a test to measure his/her reading ability. We are interested in calculating a 95% confidence
interval for the population mean score. In the sample, the mean score is 64 points, and the standard
deviation is 12 points. What is the margin of error associated with the confidence interval?
Because the standard deviation we have is s, not σ, we use t*… (with 19 df):
m = 2.093 * 12/√20 = 5.6161 points
4) Scores on the SAT Mathematics test are believed to be Normally distributed. The scores of a simple
random sample of five students who recently took the exam are 550, 620, 710, 520, and 480. What is a
95% confidence interval for μ, the population mean score on the SAT Math test?
(463.4, 688.6) points
We estimate with 95% confidence that the mean score on the SAT Math test (for this population) is
between 463.4 and 688.6 points.
5) A simple random sample of 100 postal employees is used to test if the average time postal employees
have worked for the postal service has changed from the value of 7.5 years recorded 20 years ago. The
sample mean was
= 7 years with a standard deviation of s = 2 years. Assume the distribution of the
time the employees have worked for the postal service is approximately Normal. The hypotheses
being tested are H0: μ = 7.5, Ha: μ ≠ 7.5. A one-sample t test will be used.
If we let α = 0.05, will we reject Ho and support Ha? Why or why not? (include a sketch)
Using 80 df (we really have 99 df, but that’s not in Table D), the two values that get us to our rejection
regions are ±1.990. Our calculated t statistic of – 2.5 is in the rejection region; therefore we have
enough evidence to say that the average time postal employees have worked for the postal service
HAS changed – and is now less than 7.5 years.
Suppose the mean and standard deviation obtained were based on a sample of size n = 25 postal
workers rather than 100. What do we know about the value of the P-value?

It would be larger.

It would be smaller.

It would be unchanged, because the difference between
and the hypothesized value μ = 7.5
is unchanged.

It would be unchanged, because the variability measured by the standard deviation stays the
same.
If n were smaller, then the standard error will be larger, and that will make the t-statistic smaller, too
(closer to zero). This will INCREASE the p-value of the test.
6) Bags of a certain brand of tortilla chips claim to have a net weight of 14 oz. Net weights actually vary
slightly from bag to bag. Assume net weights are Normally distributed. A representative of a
consumer advocate group wishes to see if there is any evidence that the mean net weight is less than
advertised and so intends to test the hypotheses H0: μ = 14, Ha: μ < 14. To do this, he selects 16 bags of
tortilla chips of this brand at random and determines the net weight of each. He finds a sample mean of
13.88 oz with a standard deviation of s = 0.24 oz.
What is the value of the test statistic?
Determine which of the following statements regarding the decision the representative would make is
true. The p-value of this test is between 0.025 and 0.05, so the second statement is true.

He would not reject H0 at a significance level of 0.05.

He would reject H0 at a significance level 0.05, but not at 0.025.

He would reject H0 at a significance level 0.025, but not at 0.01.

He would reject H0 at a significance level 0.01.
7) Do students tend to improve their SAT Mathematics (SAT-M) score the second time they take the
test? A random sample of four students who took the test twice received the following scores:
Student
1
2
3
4
First score
450
520
720
600
Second score
440
600
720
630
Difference
– 10
80
0
30
Assume that the change in SAT-M score (second score – first score) for the population of all students
taking the test twice is Normally distributed.
Which of the following gives the 90% confidence interval for μ, the mean change in SAT-M score?
(–186.4, 236.4)
(–140.7, 190.7)
(–22.55, 72.55)
(–39.31, 89.31)
This is paired data, so you have to analyze the differences. The mean is 25 and the standard deviation
s = 40.4145. t* = 2.353… margin of error = 47.55.
8) A study was to be undertaken to determine if a particular training program would improve physical
fitness. An SRS of 31 university students was selected to be enrolled in the fitness program. One
important measure of fitness is maximum oxygen uptake. Measurements of oxygen uptake in
untrained individuals are known to follow a Normal distribution with a mean of μ = 45 ml/kg/min. The
researchers wished to determine if there was evidence that their sample of students differed from the
general population of untrained subjects. The measurements made on the subjects coming into this
study produced a mean of
= 47.4 with a standard deviation of s = 5.3.
If we let α = 0.05, will we reject Ho and support Ha? Why or why not? (include a sketch)
t = 2.521
With 30 df, the values of t that get us to the rejection regions are ± 2.042. We are
inside the rejection region, and so we have enough evidence to suggest that the students differ from the
general population of untrained subjects.
If the 98% confidence interval were determined to be (45.2, 49.6) (an interval, which may or not be
correct), what would the researchers conclude?

The sample is not statistically significant at the 0.02 level.

The sample mean is significantly different from μ = 45 at the 0.04 level.

There would be no reason to believe the selected students come from a different population.

This suggests, at the 0.02 level of significance, that there is evidence to conclude that the
sample comes from a population with a mean different from μ = 45.
While the second statement is technically true (45 is outside the CI, so we know that p < 0.02 and
so it must also be < 0.04), the best answer is the fourth statement.
If the level of confidence were changed to 95%, what would happen to the confidence interval and the
P-value?

The confidence interval would become longer and the P-value would decrease.

The confidence interval would become shorter and the P-value would increase.

The confidence interval would not change and the P-value also would not change.

The confidence interval would become shorter and the P-value would decrease.

The confidence interval would become shorter but the P-value would not change.
Lowering the level of confidence (and hence alpha) will make the interval itself shorter
(smaller margin of error) but will have no effect on the p-value of the test, because the test
statistic t will not change.