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33 . The screening test has a 20 percent probability of a Type II error, meaning that 20 percent of the time, it will fail to detect TB when it is in fact present.
34 . Eighty percent of the time, the screening test will detect TB when it is actually present.
35 . The Student’s t -test.
36 . The normal distribution or z -test.
37 . The normal distribution with μ = p and σ = $\sqrt{\frac{pq}{n}}$
38 . t _{24} . You use the t -distribution because you don’t know the population standard deviation, and the degrees of freedom are 24 because df = n – 1.
39 .
$\overline{X}~N\left(0.95,\frac{0.051}{\sqrt{100}}\right)$
Because you know the population standard deviation, and have a large sample, you can use the normal distribution.
40 . Fail to reject the null hypothesis, because α ≤ p
41 . Reject the null hypothesis, because α ≥ p .
42 .
H
_{0} :
μ ≥ 29.0”
H
_{a} :
μ <29.0”
43 . t _{19} . Because you do not know the population standard deviation, use the t -distribution. The degrees of freedom are 19, because df = n – 1.
44 . The test statistic is –4.4721 and the p -value is 0.00013 using the calculator function TTEST.
45 . With α = 0.05, reject the null hypothesis.
46 . With α = 0.05, the p -value is almost zero using the calculator function TTEST so reject the null hypothesis.
47 . The level of significance is five percent.
48 . two-tailed
49 . one-tailed
50 .
H
_{0} :
p = 0.8
H
_{a} :
p ≠ 0.8
51 . You will use the normal test for a single population proportion because np and nq are both greater than five.
52 . They are matched (paired), because you interviewed married couples.
53 . They are independent, because participants were assigned at random to the groups.
54 . They are matched (paired), because you collected data twice from each individual.
55 .
$d=\frac{{\overline{x}}_{1}-{\overline{x}}_{2}}{{s}_{pooled}}=\frac{4.8-4.2}{1.6}=0.375$
This is a small effect size, because 0.375 falls between Cohen’s small (0.2) and medium (0.5) effect sizes.
56 .
$d=\frac{{\overline{x}}_{1}-{\overline{x}}_{2}}{{s}_{pooled}}=\frac{5.2-4.2}{1.6}=0.625$
The effect size is 0.625. By Cohen’s standard, this is a medium effect size, because it falls between the medium (0.5) and large (0.8) effect sizes.
57 . p -value<0.01.
58 . You will only reject the null hypothesis if you get a value significantly below the hypothesized mean of 110.
59 . ${\overline{X}}_{1}-{\overline{X}}_{2}$ , i.e., the mean difference in amount spent on textbooks for the two groups.
60 .
H
_{0} :
${\overline{X}}_{1}-{\overline{X}}_{2}$ ≤ 0
H
_{a} :
${\overline{X}}_{1}-{\overline{X}}_{2}$ >0
This could also be written as:
H
_{0} :
${\overline{X}}_{1}\le {\overline{X}}_{2}$
H
_{a} :
${\overline{X}}_{1}>{\overline{X}}_{2}$
61 . Using the calculator function 2-SampTtest, reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students.
62 . Using the calculator function 2-SampTtest, reject the null hypothesis. At the 1% significance level, there is sufficient evidence to conclude that the science students spend more on textbooks than the humanities students.
63 .
H
_{0} :
p
_{A} =
p
_{B}
H
_{a} :
p
_{A} ≠
p
_{B}
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