Download 8.25 Hypothesis Testing: Normal Theory 8.26 Comparing experiments

8.25. HYPOTHESIS TESTING: NORMAL THEORY
8.25
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Hypothesis Testing: Normal Theory
√
8.5.2. The sample mean is 40.26. The standard error is 4.0/ 10 = 1.265. The difference is 2.26/1.265 =
1.787 standard errors above the mean. With a one-tailed test, the significance is 0.037. With a twotailed test, it is 0.074.
p
8.5.6. The sample mean is 40.26 and the sample variance is 9.396. The standard error is 9.396/10 =
0.969. The value of the t statistic is
t=
40.26 − 38.0
= 2.332.
0.969
This just barely exceeds the critical value 2.262 of the t distribution with ν = 9 degrees of freedom, so
the result is significant.
8.5.10. This is at 2.576 standard errors above the mean, or 41.26.
8.5.14. We need the probability that the mean height is greater than 41.26 or less than 34.7 (the other
results more extreme than observed), which works out to 0.915.
8.5.26. p̂ = 0.6, Under the null hypothesis, s2 = 0.45·0.55 = 0.2475, s = 0.497, and the standard error
√
is s/ n = 0.497/10 = 0.0497. The value 0.6 is 0.15 away from the null, which is 3.02 standard errors.
With a two-tailed test, the p-value is 1 − Φ(3.02) + Φ(−3.02) = 0.0025, which is highly significant.
8.5.30. The null hypothesis has the normal approximation X ∼ N (35.0, 35.0). Using the continuity
correction and a one-tailed test,
√ − 35 ) = Pr(Z ≤ −2.11) = 0.017.
Pr(X ≤ 22.5) = Pr(Z ≤ 22.5
35
The difference is significant.
√
8.5.32. It would have to be at least 2.326 standard deviations below 35, or 35 − 2.326 35 = 21.2.
There would have to be 21 or fewer errors.
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Comparing experiments
8.6.6. Because the two sample sizes are equal, the pooled variance is the ordinary average of the
sample variances, or s2p = 3.3.
8.6.10. The standard error of the difference of the means is
r
sX̄1 −X̄2 =
3.3 3.3
+
= 0.812.
10
10
8.6.14. Using the standard error from exercise 8.6.10,
t=
3.8 − 2.1
= 2.093.
0.812
This is just barely less than the critical value 2.101 of the t distribution with n1 +n2 −2 = 10+10−2 = 18
degrees of freedom. The difference is nearly significant.
8.6.18. The distribution of the sample mean from the experimental plot is N (µ1 , 1.6) and for the
control is N (µ2 , 1.6). The distribution of the difference D is D ∼ N (µ1 − µ2 , 3.2). The difference in
sample means is 3.76, so we must find the probability that D ≥ 3.76 or D ≤ −3.76 for a two-tailed
test. Transforming into a standard normal distribution, the two-tailed test gives
D
3.76
D
−3.76
Pr(D ≥ 3.76 or D ≤ −3.76) = Pr( √
≥√
or √
≤ √
)
3.2
3.2
3.2
3.2
= Pr(Z ≥ 2.10) + Pr(Z ≤ −2.10)
= 1 − Φ(2.10) + Φ(−2.10) = 0.036.
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The difference is significant with a two-tailed test, and is less than the p-value of 0.074 found in
exercise 8.5.2. Even though we are uncertain about the true mean of the control population, our
observed mean is less than the value of 38.0 used in exercise 8.5.2, creating a larger, and significant,
difference.
8.6.22. The pooled variance, in this case with equal sample sizes, is the average of the sample variances
or
9.396 + 17.2
s2p =
= 13.30.
2
Let Hc represent the heights of the control plants. The standard error of the difference in means is
r
sH̄−H̄c =
13.30 13.30
+
= 1.63.
10
10
The observed difference in means is 3.76, so
3.76
= 2.31.
1.63
This exceeds the critical value of the t distribution with 10 + 10 − 2 = 18 degrees of freedom, producing
a significant result, as in the earlier case.
8.6.38. To compare the treatment with the control, we estimate the variance of the treatment as
0.4 · 0.6/100 = 0.0024 and the variance of the control as 0.5 · 0.5/100 = 0.0025. The variance of the
difference is then 0.0049. We are testing the probability that this is greater than 0.6 − 0.5 = 0.1.
The standard deviation is 0.07 so they differ by 1.43 standard deviations, leading to a p-value of
1−Φ(1.43)+Φ(−1.43) = 0.15. To test against the expectation of equal fraction, we find the probability
of 40 or fewer or 60 or more with p = 0.5 to be 0.046. This p-value is much smaller and is significant
because there is no uncertainty associated with the control in this case.
8.6.42. We find that s21 = 1.473 and s22 = 3.684. Because the two sample sizes are equal, the pooled
variance is the ordinary average of these values, so
t=
s2p =
s21 + s22
= 2.579.
2
The standard error of the difference is
r
sX̄1 −X̄2 =
2.579 2.579
+
= 0.718
10
10
The sample means are X̄1 = 5.488 and X̄2 = 6.592. Therefore
X̄1 − X̄2
sX̄1 −X̄2
5.488 − 6.592
=
= −1.538.
0.718
The number of degrees of freedom is ν = n1 + n2 − 2 = 18. The value of t is less than the critical value
2.101 from table 8.3.4, so there is no significant effect of this medication.
8.6.44. The differences are 2.313, 1.478, 1.476, 1.127, −0.017, 1.146, −0.164, 2.281, 1.557, −0.158. These
values have sample variance s2Y = 0.866, and thus standard error
t =
r
sY
= 0.294.
10
Our null hypothesis is that the mean of Y is equal to zero, while the actual sample mean of Y is 1.104.
The t statistic is
1.104
t=
= 3.755.
0.294
This value exceeds the critical value for a p-value of 0.01 with 9 degrees of freedom, a stronger result
than found with the unpaired test.
sȲ =
8.27. ANALYSIS OF CONTINGENCY TABLES AND GOODNESS OF FIT
8.27
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Analysis of contingency tables and goodness of fit
8.7.18. To the expected numbers if students attend independently, we must find the fraction in each
column. There are a total of 182 days, student 1 shows up on 125 of them, for a fraction of 0.687.
Multiplying by the number of days that student 2 attends or not (145 and 37) gives
student 2 attends
student 2 does not attend
student 1 attends
99.615
25.419
student 1 does not attend
45.385
11.581
Then
χ2 = 6.847.
There is 1 degree of freedom in this 2 by 2 table, and this value slightly exceeds the critical value for
p=0.01. These students seem to be avoiding each other.
8.7.32. We have 21 plants expressing the recessive trait and 99 that do not. We expect 30 to express
the recessive trait and 90 that do not. Then
χ2 =
(21 − 30)2 (99 − 90)2
+
= 3.6.
30
90
This is slightly less than the critical value with 1 degree of freedom, we cannot reject the hypothesis of
Mendelian ratio.
8.7.40. The table of outcomes is
male second
female second
male first
16
3
female first
11
14
The fraction of males among the first offspring is 19
44 = 0.432. The expected numbers are then
male second
female second
male first
11.66
7.34
female first
15.34
9.66
Then χ2 = 7.359. This is now higher than the value found in Example 8.7.15, and is still significant
at the p=0.01 level. The increased value is due to the fact that in broods with exactly one male, it is
more likely to have the male first.