Name: __________________________________
1.
a.
b.
c.
d.
2.
a.
b.
c.
d.
A Type I error occurs when we:
reject a false null hypothesis.
reject a true null hypothesis.
don't reject a false null hypothesis.
don't reject a true null hypothesis.
A professor of math is testing the claim that the average student spends 3 hours studying for
a midterm exam. Which hypothesis should she use to test the claim?
H0:
3, HA: = 3
H0: = 3, HA:
3
H0:
3, HA: = 3
H0: = 3, HA: < 3
When we reject the null hypothesis, we conclude that:
a. there is enough statistical evidence to infer that the alternative hypothesis is true.
b. there is not enough statistical evidence to infer that the alternative hypothesis is
true.
c. there is enough statistical evidence to infer that the null hypothesis is true.
d. the test is statistically insignificant at whatever level of significance the test was
conducted.
3.
4.
a.
b.
c.
d.
5.
a.
b.
c.
d.
The p-value of a test is the probability:
we would see a test statistic as extreme as the one we observe if the null hypothesis
were true.
of making a Type I error.
we would see a test statistic as extreme as the one we observe if the null hypothesis
were false.
of making a Type II error.
Which of the following p-values will lead us to reject the null hypothesis if
0.150
0.100
0.055
0.025
= 0.05?
1
Name: __________________________________
6.
A courier company in Atlanta wants to determine the mean time it takes his company to pickup and deliver a package anywhere within the city limits. He conducts a study of 50
deliveries and this random sample gives him an average delivery time (x-bar) of 2.8 hours
and a sample standard deviation (s) of .6 hours. Determine a confidence interval for the mean
delivery time (μ) using a 95% confidence level and explain, briefly, what this confidence
interval means.
𝑛 = 50 𝑥̅ = 2.8 𝑠 = .6 𝑑𝑓 = 49
.6
. 95 𝐶𝐼 = 2.8 ± 2.01
√50
= 2.8 ± .1706
𝟐. 𝟔𝟐𝟗𝟒 𝒕𝒐 𝟐. 𝟗𝟕𝟎𝟔
If we wanted a 90% confidence interval, would the spread of the t values in our interval be
larger or smaller; that is, would the limits be closer to or further from the estimated mean?
The interval for a 90% confidence interval would be smaller; i.e., the limits would be closer
to the estimated mean.
7.
Technical data provided with a room dehumidifier explain that units tested from the
production line consume a mean (μ) of 800 watts of power, with a standard deviation (σ) of
12 watts. The superintendent of a large office complex, upon evaluating 30 of the units
recently purchased and found their mean power consumption to be 805 watts, claims that her
shipment does not meet company specifications. At the α=0.05 level of significance, is the
superintendent’s complaint justified? Include your null and alternative hypotheses and a
brief statement of your conclusion.
𝐻0 : 𝜇 = 800
𝐻𝐴 : 𝜇 ≠ 800
I would use a two-tailed test for this one. If you use a one-tail test, I will evaluate it based on
your logic.
805 − 800
5
=
= 2.28
12⁄
2.19
√30
Critical values for a 2-tailed Z-test with a significance level of .05 = -1.96 and +1.96
𝑇𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = 𝑍 =
Since my test statistic is more extreme than my critical value, I have enough evidence to
reject the null hypothesis and conclude the alternative hypothesis must be true. The
superintendent’s claim that the shipment does not meet company specifications is justified.
2
Name: __________________________________
8.
Using traditional methods, it takes a new flight student an average (μ) of 10.5 hours of
training before he or she can solo. Assume this population is normally distributed. A new
training method combines computer aided instruction with the old methods. Using this new
method, an instructor conducted a random sample of 26 students and discovered the average
time it took this group to solo was 10.9 hours with a sample standard deviation (s) of 1.7
hours. Is there evidence at the α=.05 level of significance to conclude that the new technique
lengthens training time? Include your null and alternative hypotheses and a brief statement of
your conclusion.
Because the question asks, “Is there evidence at the α=.05 level of significance to conclude
that the new technique lengthens training time?” I would use a one-tailed test. You may have
decided on a 2-tailed test, which is OK.
𝐻0 : 𝜇 ≤ 10.5
𝐻𝐴 : 𝜇 > 10.5
We don’t know sigma, so the test statistic has to be a t.
10.9 − 10.5
= 1.2
1.7⁄
√26
The critical value for a t-statistic with 25 degrees of freedom and a 1-tailed alpha of .05 =
1.71
𝑇𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = 𝑡 =
Because our test statistic is less extreme than our critical value, we do not have sufficient
evidence to reject the null hypothesis. We cannot conclude that the new technique lengthens
training time.
9.
A study was conducted of 90 adult male patients to see if a new treatment for congestive
heart failure would increase their capacity for exercise beyond the increase they saw using
the old treatment. The old treatment produced an average increase (μ) of 2 minutes. The
study of the new treatment showed an average increase of 2.17 minutes with a sample
standard deviation (s) of 1.05. Use a significance level (α) of .10. Using the software
package, Minitab®, I conducted a one-sample t-test and got the following output:
Test of mu = 2 vs ≠ 2
N
90
Mean
2.17000
StDev
1.05000
SE Mean
0.11068
T
1.54
P
0.064
Based on the output shown above, what conclusions can we draw about the effectiveness of
the new treatment?
All we need to look at are the p-value and the significance. In this case, the p of 0.064 is less
than the alpha of .10. We, therefore, have sufficient evidence to reject the null hypothesis and
conclude that the mean of the new treatment is not equal to 2.
3
Name: __________________________________
10.
The average length of a hospital stay (μ) in community hospitals is 5.7 days. Assume a
population standard deviation (σ) of 2.5 days and a random sample of 50 patients. What is
the probability that the average length of stay for these 50 patients will be 6.5 days or less?
This is a sampling distribution and we can use z-values because we know the value of sigma.
We are also safe assuming the sampling distribution is normally distributed because n>30.
6.5 − 5.7
= 2.26
2.5⁄
√50
𝑃(𝑋 < 6.5) = .9880
𝑍=
If the sample size had been 8 patients instead of 50, what further assumption would you have
to make in order to solve this problem?
We would need to assume that the length of hospital stays was normally distributed for the
population.
11.
A corporation maintains a large fleet of cars for its salespeople. To check the average number
of miles driven per month per car, a random sample of n=40 cars is examined. The mean and
standard deviation for the sample are 𝑥̅ = 2,752 miles and s = 350 miles. Records for previous
years indicate the average number of miles driven per car per month was 2,600. Using the
data from the sample, construct a test to determine if the sample mean (2,752) is significantly
different from the historical mean (2,600). Use a significance level of .05.
Since we’re being asked if the sample is significantly different from the historical mean, and
not being asked if it’s greater than or less than, this should be a 2-way test.
𝐻0 : 𝜇 = 2600
𝐻𝐴 : 𝜇 ≠ 2600
We will have to use t-statistics because we don’t know the value of sigma.
2752 − 2600
= 2.74
350⁄
√40
For critical values, we have 39 degrees of freedom and α/2 = .025, so our critical values are
plus and minus 2.02.
𝑇𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 = 𝑡 =
Since our test statistic is more extreme than our critical value, we have sufficient evidence to
reject the null hypothesis and conclude the alternative hypothesis must be true. The sample
mean is different from the historical mean.
4
Name: __________________________________
Critical values using Z
Two-tailed
𝛼 = .10: + 𝑎𝑛𝑑 − 1.65
𝛼 = .05: + 𝑎𝑛𝑑 − 1.96
One-tailed
𝛼 = .10: + 𝑜𝑟 − 1.28
𝛼 = .05: + 𝑜𝑟 − 1.65
5