0.2 Practice tests (1-4) and final exams  (Page 19/36)

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.

9.3: distribution needed for hypothesis testing

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.

9.4: rare events, the sample, decision, and conclusion

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.

9.5: additional information and full hypothesis test examples

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.

10.1: comparing two independent population means with unknown population standard deviations

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.

10.2: comparing two independent population means with known population standard deviations

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.

10.3: comparing two independent population proportions

63 . H ₀ : p _A = p _B
H _a : p _A ≠ p _B