Download Chapter 9: Solution

Chapter 9: Solution Manual for Introductory Biological Statistics, 3rd edition, Raymond E.
Hampton and John E. Havel
Lianfen Qian ®
Chapter 9: Statistical Inference of One Population Mean
9-1:
We are given
n  30, x  70.2, s  10.51,1    .95.
From Table A.2, we have t( 0.05, 29)  2.045
Thus the 95% confidence interval for the population mean (the mean pulse rate) is
x  t( , n 1) 
s
 70.2  (2.045)(10.51/ 30 )  70.2  (2.045)(1.9189)  70.2  3.924  (66.28,74.12).
n
9-3:
We are given
n  9, x  19.256, s  0.855,1    .95 and 1    .99
From Table A.2, we have t(0.05,8)  2.306, t( 0.01,8)  3.355
Thus the 95% confidence interval for the population mea is
x  t( , n 1) 
s
 19.256  (2.306)(2.564 / 9 )  19.256  (2.306)(.855)  19.256  1.971  (17.28,21.23).
n
Thus the 99% confidence interval for the population mea is
x  t( , n 1) 
s
 19.256  (3.355)(2.564 / 9 )  19.256  (3.355)(.855)  19.256  2.867  (16.39,22.12).
n
9-5: We are testing H0: μ=100 vs. Ha: μ100 grams
We are given
n  20, x  97, s  2.5
The computed t-test statistic is
t
x  100 97  100
3


 5.37
s/ n
2.5 / 20 0.559
Conclusion: We are almost 100% confident to conclude that the mean weight is significantly
different from 100 grams with a p-value=0.000. Hence they should be removed from the
breeding program.
9-7:
We are testing H0: μ≤3.5 vs. Ha: μ>3.5 mg per cigarette
We are given
n  10, x  4.2, s  1.4
The computed t-test statistic is
t
x  3.5 4.2  3.5
0.7


 1.581 which is between 1.383 and 1.833 for df=9. The
s / n 1.4 / 10 0.443
corresponding two-tailed probabilities are 0.2 and 0.1. Since we are testing one-tailed
alternative hypothesis, so the one one-tail probability is between 0.1 and 0.05. That is,
0.05<p-value<0.1. Hence we fail to reject the null hypothesis at 0.05 significance level and
conclude that there is not strong enough evidence to reject the manufacturer’s claim.
9-9: We are testing H0: μ≥50 vs. Ha: μ<50 grams
We are given
n  20, x  46.1, s  6.025
The computed t-test statistic is
x  50
46.1  50
 3.9


 2.895 which is between -2.861 and -3.883 from the table
s / n 6.025 / 20 1.347
(with a negative sign added) for left-tailed alternative hypothesis) with df=19. Thus, the p-value
(the left tail probability) is between 0.0005 and 0.005 leading to conclude that there is strong
enough evidence to reject the null hypothesis. That is, the enzyme activities are depressed in
this population and hence there is significant indication of the presence of the pollutant.
t
9-11. We are testing H0: μ≥8.0 vs. Ha: μ<8.0 ppm
We are given
n  20, x  7.94, s  0.204
The computed t-test statistic is
x  8.0
7.94  8.0
 0.06


 1.3176 which is bigger than -1.328 from the table for lefts / n 0.204 / 20 0.0455
tailed alternative hypothesis ( a negative sign added) with df=19. Thus, the p-value (the left tail
probability) is bigger than 0.2/2=0.1 leading to conclude that there is not enough evidence to
reject the null hypothesis. That is, the average weight of the chicks is not below normal weight
of 8.0 kg.
t
9-13. We are testing H0: μd =0 vs. Ha: μd 0
We are given
n  8, xd  35.625, s  23.6156
The computed t-test statistic is
xd  0
35.625
35.625


 4.267 which is between 3.499 and 5.408 from the table for
s / n 23.6156 / 8 8.349
two-tailed alternative hypothesis with df=7. Thus, the p-value is between 0.001 and 0.01 leading
to conclude that there is enough evidence to reject the null hypothesis. That is, the pre-exercise
and post-exercise pulse rates are significantly different
t
9-15. We are testing H0: μd ≥0 vs. Ha: μd <0
We are given
n  12, xd  3, s  1.809
The computed t-test statistic is
xd  0
3
3


 5.745 which is between -5/921 and -4.437 from the table
s / n 1.809 / 12 .522
(with negative sign added) for left-tailed alternative hypothesis with df=11. Thus, the p-value is
between 0.00005 and 0.0005 leading to conclude that there is strong evidence to reject the null
hypothesis. That is, Cortisone decrease long term memory significantly.
t
9-17. We are testing H0: md ≤0 vs. Ha: md >0
T=1.5+1.5=3, and W=78.
From A.5, n=12, c.v.=7 at α=0.005. Reject H0 at 0.005 significant level to conclude that the
vaccine increase the encounter rate significantly.
Minitab output shows that
9-19: We are given: n=8 (7(+),1(-))
Let Y=the number of subjects who reported an improvement. Then Y~Bin(8,.5) under null
hypothesis of no change. P-value=P(Y≥7)=p(7)+p(8)=p(1)+p(0)=(1+8)(.5)8=0.0352. We conclude
there is sufficient evidence of improvement with p-value=0.0352.