Stat 281: Introduction to Statistics
F2005, Dwight Galster
Homework 8b-9
Key
Problems 8.62, 8.67 and 8.68 on pages 357-358
Problems 8.86, 8.87, 8.92, and 8.94 on page 370-371
Problem 9.29 on page 405
Problem 9.31 on page 406
Problem 9.40 on page 409
Problem 9.1 on page 391
Problem 9.9 on page 392
Problem 9.18 (assume equal variances) on page 395
Problem 9.26 (assume unequal variances) on page 397
8.62
a) .69 is pretty far below .75 in proportions, so it would seem to support the alternative
hypothesis and contradict the null hypothesis. (Answer may vary)
b)
1) Ho: p=.75 Ha: p<.75
2) z-test for a proportion, check assumptions: np=57>5, nq=25>5, n=100>30
3) Classic approach: Reject Ho if z*<-1.645, p-value approach: Reject Ho if p<.05.
4) z*=(.69-.75)/sqrt(.75*.25/100)=-1.3856, p=.0823
5) Do not reject Ho and conclude that there is not sufficient evidence to show p<.75.
This does not agree with my intuition, and I am surprised that the test did not result in a
rejection. (Answer may vary)
c) The p-value is .0823. It means that for any alpha of more than .0823 we would reject
Ho. Since alpha is .05 in this problem, we do not reject Ho.
8.67
1) Ho: p=.45, Ha: p>.45
2) z-test for a proportion, check assumptions: np=375.75>5, nq=459.25>5, n=835>30.
3) Classic approach: Reject Ho if z*>1.645, p-value approach: Reject Ho if p<.05.
4) z*=(.48-.45)/sqrt(.45*.55/835)=1.74, p-value=.0409.
5) We reject Ho and conclude that more than 45% of males are raised in single parent
families.
8.68
1) Ho: p=.6, Pa: p>.6
2) z-test for a proportion, check assumptions: np=19.8>5, nq=13.2>5, n=33>30.
3) Classic approach: Reject Ho if z*>1.645, p-value approach: Reject Ho if p<.05.
4) z*=(.697-.6)/sqrt(.6*.4/33)=1.14, p-value=.1271.
5) Do not reject Ho and conclude that we have no evidence that more than 60% of
women over age 40 will improve their skin by using Pond’s Age-Defying Complex.
8.86
a) 17.275
b) 15.5073
c) 11.1433
8.87 [I use the symbol X* for “chi-square-star” here for ease of typing]
a) Reject Ho if X*<6.26214 or X*>27.4884
b) Reject Ho if X*>38.9321
c) Reject Ho if X*>21.0642
8.92
1) Ho: sigma-squared=200,000, Ha: sigma-squared<200,000
2) X* test for a variance, Assume population is at least approximately normal
3) Reject Ho if X*<5.22935 (or p<.01)
4) X*=15(410^2)/200,000=12.6075, p=.3774
5) Do not reject Ho and conclude there is no evidence that the variance of birthweights
of babies born to cocaine-dependent women is less than 200,000.
8.94
1) Ho: sigma-squared=.00156, Ha: sigma-squared>.00156
2) X* test for a variance, Assume population is at least approximately normal
3) Reject Ho if X*>123.225, (or p<.05) Using an assumed alpha of .05 (not stated)
4) X*=133.269, p=.0123.
5) Reject Ho and conclude that the variance of the diameters under the new process is
greater than the variance under the old process.
9.29
a) Differences are -3, -2, -2, -4, 0, -1. d-bar (or x-bar-D) is -2, and s-squared-d (D) is 2.
b) mu-D=mu-2 – mu-1.
c) d-bar +- t(5, .025) * s-d-bar, -2 +- 2.571(.57735), -2 +- 1.484, (-3.484, -0.516)
d)
1) Ho: mu-d = 0, Ha: mu-d <> 0.
2) Paired difference t test. check assumptions: sigma unknown, Assume both
populations are normal so that differences are normal and therefore d-bar is normal.
3) Reject Ho if t*<-2.571 or t*>2.571 (or p<.05)
4) t*= -2/.57735=-3.4641, p=.017
5) Reject Ho and conclude that the means difference of these populations is not zero.
9.31
a)
1) Ho: mu-d=10, Ha: mu-d <>10
2) Paired difference t test, sigma unknown, Assume the populations are normal.
3) Reject Ho if t*<-2.447 or t*>2.447 (or p<.05)
4) t*=(11.8571-10)/(6.067/sqrt(7))=.81, p=.449
5) Do not reject Ho, and conclude that the mean of the differences between the
populations may be 10.
b) Reject Ho for alpha greater than .449. In other words, for any reasonable alpha value,
this test will not result in a rejection.
9.40 [use the information from the computer output]
1) Ho: mu-1 = mu-2, Ha: mu-1 <> mu-2
2) Paired difference t test, sigma unknown, Assume the populations are normal.
3) (No need to state classic approach rejection region) Reject Ho if p<.10.
4) From the output, p=.653
5) Do not reject Ho and conclude that there is no evidence that the male and female pups
differ in the number of swims required to complete the maze 3 times.
9.1
a) 35 +- 1.96(sqrt(150^2/400+200^2/400)), (10.5, 59.5)
b)
1) Ho: (mu-1 – mu-2)=0, Ha: (mu-1 – mu-2) <> 0
2) t-test for difference of means, independent samples, no reason to believe variances are
equal, samples large enough to ensure normality.
3) Reject Ho if p<alpha, unspecified (use .05)
4) t*=35/12.5=2.8, df=399 (off the chart, use z approximation), p=.0052
5) Reject Ho and conclude that the means of the two populations differ.
c) The only difference is the p-value would be .0026, still resulting in a rejection.
d)
1) Ho: (mu-1 – mu-2)=25, Ha: (mu-1 – mu-2) <> 25
2) t-test for difference of means, independent samples, no reason to believe variances are
equal, samples large enough to ensure normality.
3) Reject Ho if p<alpha, unspecified (use .05)
4) t*=10/12.5=0.8, df=399 (off the chart, use z approximation), p=.4238
5) Do not reject Ho and conclude that the difference between the means of the two
populations may be 25.
Since the sample gave a difference of 35, we can see that a null hypothesis difference of
25 is much closer to that than 0, so naturally there would not be as much evidence to
reject this hypothesis, and the results of the two test shows that is indeed the case.
e) these are given in step 2 of the hypothesis test.
9.9 [using computer output]
a) The p-value of the test is .115 Since alpha is unspecified, we will use .05. We fail to
reject Ho and conclude that there is no evidence that the two population means differ.
b) The p-value would be half of the previous, or .0575. At alpha =.05 the conclusion is
the same.
9.18 [assume equal variance]
1) Ho: (mu-1 – mu-2)=0, Ha: (mu-1 – mu-2) <> 0
2) t-test for difference of means, independent samples, variances are believed to be
equal, populations must be approximately normal to conduct the test, because samples are
small. Use pooled variance formula.
3) Degrees of freedom for pooled variance is 47. Reject Ho if t*<-2.01 or t*>2.01, or if
p<.05
4) pooled variance is .0081, t*=.04/.02572=1.555, p=.127
5) Do not reject Ho and conclude that there may be no difference between the two types
of castration with regard to the number of high frequency vocal responses.
9.26 [assume unequal variance]
1) Denote the green blades as the first sample and the decayed blades as the second
sample. The null hypothesis of no difference in feeding rates is then Ho: (mu-1 – mu2)=0, and the alternative hypothesis is that the decayed blades are eaten more quickly, so
we designate Ha: (mu-1 – mu-2) > 0
2) t-test for difference of means, independent samples, variances are not believed to be
equal, populations must be approximately normal to conduct the test, because samples are
small. Use non-pooled variance formula.
3) Degrees of freedom is 9. Alpha is unspecified, so use .05. Reject Ho if t*>1.833, or if
p<.05
4) t*=.99/.355=2.789, p=.01
5) Reject Ho and conclude that the sea urchins eat the decayed blades faster than the
green blades, and therefore probably find them more palatable.