STAT 421
SOLUTIONS TO EXAM 2 EXTRA PRACTICE PROBLEMS
Problem 13.25 – see solution in back of text.
Problem 13.36
The first step is to set up the hypotheses and the α level. Here, we have α = 0.05 and
H0 : µ1 − µ2 = 0 versus H1 : µ1 − µ2 6= 0.
The second step is to determine the appropriate critical region. Here, we’ll compute a
z-statistic and reject H0 when |z| > zα/2 = 1.96.
The third step is to compute the value of the test statistic for the data,
9.1 − 8.0
z=q
= 2.59.
2.12
1.92
+ 50
40
The final step is to make a statement concerning the null hypothesis. Since z = 2.59 > 1.96 =
zα/2 , we reject H0 . We conclude that the mean number of lunches claimed as deductible
expenses per month is significantly different for the banking and insurance industries.
Problem 13.37
Referring to the previous problem, we find that the p-value is p = P (|Z| > 2.59) = 2∗P (Z >
2.59) = 2 ∗ (0.5 − 0.4952) = 0.0096. Since p < 0.05 = α, we reject H0 .
Problem 13.40
Although the sample sizes are small for both samples, we can assume that both populations
are normal and that their variances are equal, and so we can apply the t-test to test H0 :
µ1 − µ2 = 0 versus H1 : µ1 − µ2 6= 0 at level α = 0.01. To carry out the test, we’ll compare
the t statistic computed from the data to t0.005,10 = 3.169. Next, we compute the pooled
sample variance
(n1 − 1)s21 + (n2 − 1)s22
s2p =
= 7.65
n1 + n2 − 2
and the test statistic
77.4 − 72.2
t= q
= 3.26
7.65( 16 + 61 )
Since t = 3.26 > 3.169, we reject H0 and conclude that there is a significant difference in the
mean cephalic index for inhabitants of the two Pacific islands.