Exam-style Questions (not intended to be comprehensive) - Solutions
1) A researcher is testing the claim that the mean income of male executives is higher
than the mean income of female executives. Write the null and alternative hypothesis
in symbolic form, and indicate which represents the original claim.
H0: mu1 = mu2
H1: mu1 > mu2
“Claim” (mu1 is mean male income, mu2 is females)
2) A pollster wants to test whether the proportion of people supporting a political
candidate is greater than 50%. this test is right tailed
3) Testing the claim H1: μ < 21 with a 0.05 significance level and sample size 39, find
the critical value.
Left-tailed test. Since it’s mu, it’s a means test, so use a t-value for critical value. n is
39, so degrees of freedom is 38. Area in one tail is 0.05. Value from table is 1.686.
Since it’s left-tailed, the critical value is negative, so the critical value is t = -1.686.
4) Testing the claim H1: p < 0.40, you calculate your test statistic to be z = -1.84.
Calculate the P-value for this test.
Left tailed test, so P-value is area to left of test statistic. We need P(z < -1.84). Using the
z-table, the area is 0.0329, so the P-value of this test is 0.0329
5) In a right-tailed test, you calculate the critical value to be 1.645, and your test statistic
to be 1.34. This would lead you to FAIL TO REJECT the null hypothesis.
6) Testing the claim H1: p < 0.40, you calculate your test statistic to be z = -1.84, and the
critical value to be –2.33. Sketch a graph of the distribution, labeling the test statistic
and critical value, and shading the critical region.
7) In a survey of 500 people, 160 reported having been on a diet in the last year. Use
this sample to test the claim that more than 30% of people have been on a diet in the
last year. Test at the 10% significance level.
Since our claim is about a percentage of people, this is a test of proportions.
Claim: p > 0.30
H0: p = 0.30
H1: p > 0.30 “Claim”
Alpha = 0.10
p-hat = 160/500 = 0.32
Test stat: z = (.32 - .30)/sqrt(.30*.70/500) = 0.976
Critical value: Since this is a right-tailed test, 0.10 is above the critical value, 0.90 below.
Looking at the z-table, the z-value that gives an area below of .90 is 1.28
Since the test statistic does not fall in the critical region, we fail to reject the null
hypothesis. Conclusion: There is not sufficient sample evidence to support the claim
that more than 30% of people have been on a diet in the last year.
8) A quality control agent randomly takes 56 bags of M&Ms off the production line.
She finds the mean number of M&Ms in each bag to be 56.8, with a standard
deviation of 1.44. Test the claim that the overall mean number of M&Ms in each bag
is at least 55. Test at the 5% significance level.
The claim is about a mean, so this is a means test. Claim: mu > 55
H0: mu = 55
H1: mu > 55 “Claim”
Alpha = 0.05
x-bar = 56.8, s = 1.44, n = 56
Test stat: t = (56.8 – 55) / (1.44/sqrt(56)) = 9.35
Critical t: Right tailed test, so 0.05 area in one tail, 55 degrees of freedom. From the ttable, the t critical value is 1.673
The test statistic is huge compared to the critical value, so the test statistic does fall in the
critical region, leading us to reject the null hypothesis.
Conclusion: The sample data support the claim that the mean number of M&Ms in each
bag is at least 55.
Math 281 – Chapter 8
1. In an experiment, subjects are briefly flashed different images in each eye, one
violent and one non-violent. The researchers want to test the claim that police
officers will identify more of the images as violent than people are not police officers.
To test this claim, they perform the experiment on a group of police officers and a
group of people who are not police officers. The police officers identified 26.2% of
the 800 images shown as violent. The comparison group identified 23.7% of the 600
images shown as violent.
a. Perform a P-value method hypothesis test with a significance level of 0.01,
clearly detailing each step, and writing the conclusion out in words.
Claim: p1 > p2 (p1 is police officers, p2 is non-officers)
H0: p1 = p2
H1: p1 > p2
Alpha = 0.01
p1-hat = .262, p2-hat = .237
n1 = 800, n2 = 600
x1 = 800*.262 = 210, x2 = 142
p-bar = (210+142)/(800+600) = .251
q-bar = 1-.251 = .749
test stat:
z = [(.262 - .237) – 0 ] / sqrt(.251*.749/800 + .251*.749/600) = 1.068
P-value: Since it is a right tailed test, the p-value will be the area to the right of the test
statistic: P(z > 1.068)
P(z > 1.068) = 1 – P(z < 1.068) = 1 - .8577 = 0.1423
Since the P-value of 0.1423 is larger than the significance level of .01, we fail to reject
the null hypothesis; there is not sufficient sample evidence to support the claim that
police officers will identify more of the images as violent than people are not police
officers.
b. Find the 95% confidence interval for the difference of the population
proportions.
For the 95% confidence interval, z-critical is 1.96
E = 1.96*sqrt(.262*.738/800 + .237*.763/600) = 0.046
p1-hat - p2-hat = 0.262 – 0.237 = 0.025
0.025 – 0.045 < (p1 - p2) < 0.025 + 0.045
-0.021 < (p1 - p2) < 0.071
2. A group of students last winter decided to compare the ages of male and female
Pierce College students. The used the campus data analyst to obtain a sample of 1082
students from each population. For men, the mean age was 27, with a standard
deviation of 9.82. For women, the mean age was 29, with a standard deviation of
11.52. Test the claim that the mean age of male Pierce College students is lower than
the mean age of female students. Test at the 5% significance level.
Claim: mum < muf
(mean of males age is less than mean of females age)
H0: mum = muf
H1: mum < muf “Claim”
x-barm = 27, sm = 9.82, nm = 1082
x-barf = 29, sf = 11.52, nf = 1082
Test stat:
t = [ (27 – 29) – 0] / sqrt(9.822/1082 + 11.522/1082) = -4.35
Critical value: Left tailed test, 1081 degrees of freedom, area in one tail of 0.05. Critical
value is -1.646
Since the test statistic falls in the critical region, we reject the null hypothesis.
The sample data support the claim that the mean age of male Pierce College students is
lower than the mean age of female students.
3. Test whether there is a correlation between the weight of a car (x) and its price (y).
Using 30 sample data, your calculator readout looks like this:
LinReg
y=ax+b
a=13.0583854
b=-19955.26483
r2=.8685776
r=.9319751091
a) Does the sample data suggest a significant correlation between weight and
price of a car (at 0.01 significance level)? (include the critical value)
Critical value: 0.463
The sample data does support the claim of significant correlation
b) Use the regression to predict a car’s price if it weights 3000 pounds.
y = 13.0583854*3000-19955.26 = $19,220