8.8 Review Questions
 Determine whether the following interpretations of a 95% confidence interval (337, 343) ml for the population mean volume of beer [latex]\mu[/latex] are true or false. If false, correct it.
 We can be 95% confident that [latex]\mu[/latex] is somewhere between 337 ml and 343 ml.
 We can be 95% confident that the sample mean [latex]\bar x[/latex] is somewhere between 337 and 343 ml.
 The probability that the population mean [latex]\mu[/latex] is within the interval (337, 343) is 0.95.
 The probability that the sample mean [latex]\bar x[/latex] is within the interval (337, 343) is 0.95.
 95% of the [latex]\bar x[/latex] values are within the interval (337, 343).
 Determine whether the following statements about the [latex]P[/latex]value are true or false. If false, correct it.
 We should reject the null hypothesis [latex]H_0[/latex] if the [latex]P[/latex]value[latex]\le \alpha[/latex].
 We should accept the null [latex]H_0[/latex] if the [latex]P[/latex]value[latex]> \alpha[/latex].
 [latex]P[/latex]value is the probability that the null [latex]H_0[/latex] is true.
 [latex]P[/latex]value is the probability of rejecting [latex]H_0[/latex].
 Suppose you perform a statistical test to decide whether a nuclear reactor should be approved. Further, suppose that failing to reject the null hypothesis (the reactor is safe to use) corresponds to approval.
 Write down the null and alternative hypotheses.
 What are the type I and type II errors in this example?
 Which error has more serious consequence, type I or type II? Would you like to set [latex]\alpha[/latex] or [latex]\beta[/latex] to be relatively small?
 The mean retail price of agriculture books in 2005 was $57.61. This year’s retail mean price for 28 randomly selected agriculture books was $54.97. Assume that the population standard deviation of prices for this year’s agriculture books is $8.45.
 At the 10% significance level, do the data provide sufficient evidence to conclude that this year’s mean retail price of agriculture books has changed from the 2005 mean?
 What is the [latex]P[/latex]value of the test in part (a)?
 Obtain a confidence interval corresponding to the test in part (a).
 Does the interval obtained in part (c) support the result in part (a)?
 The ankle brachial index (ABI) compares the blood pressure of a patient’s arm to the blood pressure of the patient’s leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. Researchers obtained the ABI of 100 women with peripheral arterial disease and obtained a mean ABI of 0.64 with a standard deviation of 0.15.
 At the 5% significance level, do the data provide sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI?
 What is the [latex]P[/latex]value of the test in part (a)?
 Obtain a confidence interval corresponding to the test in part (a).
 Does the interval obtained in part (c) support the conclusion in part (a)?
Show/Hide Answer

 True, a standard way to interpret the confidence interval.
 False. The sample mean [latex]\bar{x}[/latex] is the center of the interval; we should be 100% confident that the sample mean [latex]\bar{x}[/latex] is in the interval.
 False. There is no randomness here and hence there is no probability, since the population mean [latex]\mu[/latex] is a constant and the interval (337, 343) is also fixed. [latex]\mu[/latex] is either within the interval or outside the interval.
 False. Similar arguments as the previous. The sample mean [latex]\bar{\mu}[/latex] is a fixed number, and the interval is also fixed; there is no randomness.
 False. 95% of the [latex]\bar{\mu}[/latex] values are within the interval [latex]( \mu  1.96 \frac{\sigma}{\sqrt{n}}, \mu + 1.96 \frac{\sigma}{\sqrt{n}} ).[/latex]

 True.
 False. In general, never accept [latex]H_0[/latex].
 False. Pvalue measures the strength of the evidence that the data contradicts [latex]H_0[/latex] and is in favour of [latex]H_a[/latex].
 False. Pvalue is the probability of observing [latex]z_0 (t_0)[/latex] or more extreme values.

 [latex]H_0[/latex]: the nuclear reactor is safe versus [latex]H_a[/latex]: the nuclear reactor is not safe.
 Type I error: disapprove of the nuclear reactor of ruse given that the nuclear reactor is actually safe.
Type II error: approve the nuclear reactor for use given that the nuclear reactor is not safe.  Type II error is more severe than type I. We probably need to set the type II error rate relatively small.

 Assumptions:
We have a simple random sample.
We have a large sample with [latex]n=100 >30.[/latex]
Population standard deviation [latex]\sigma[/latex] is unknown.
We can use a onesample ttest. Summarize the information: [latex]n =100, \bar{x} = 0.64, s= 0.15[/latex]. The six steps to perform a onesample ttest are: Hypotheses: [latex]H_0: \mu \ge 0.9 \text{ versus } H_a : \mu < 0.9.[/latex]
 The significance level [latex]\alpha = 0.05.[/latex]
 Observed test statistic:
[latex]t_o = \frac{\bar{x}  \mu_0}{s / \sqrt{n}} = \frac{0.64  0.9}{ 0.15 / \sqrt{100}} = 17,333[/latex]
with [latex]df = n 1 = 99.[/latex]  A lefttailed test, Pvalue = [latex]P( t \le t_0 ) = P(t \le 17.333 ) = P(t \ge 17.333 ) < 0.005[/latex].
 Since Pvalue < 0.005 < 0.05[latex](\alpha)[/latex], we reject [latex]H_0[/latex].
 At the 5% significance level, we have sufficient evidence that, on average, women with peripheral arterial disease have an unhealthy ABI
 Pvalue < 0.005
 A lefttailed test at significance level [latex]\alpha = 0.05[/latex] corresponds to a [latex]( 1 \alpha ) \times 100%[/latex] lowertailed confidence interval. With df = 99 not given in Table IV, use df = 90 the closed one but still no more than 99, [latex]\alpha =0.05 \Longrightarrow t_{\alpha} = t_{0.05} = 1.662[/latex]
[latex](\infty, \bar{x} + t_{\alpha} \frac{s}{\sqrt{n}} ) = ( \infty, 0.64 + 1.662 \times \frac{0.15}{\sqrt{100}} ) = (\infty, 0.665).[/latex]
Interpretation: we can be 95% confident that the mean ABI of women with peripheral arterial disease is somewhere below 0.665.
Note: In this course, you are only required to know how to obtain a twotailed interval. For df=99 (use 90), [latex]t_{\alpha / 2} = t_{0.025} = 1.987[/latex]. The 95% twotailed interval is
[latex]\bar{x} \pm t_{\alpha/2}\frac{s}{\sqrt{n}} = 0.64 \pm 1.987 \times \frac{0.15}{\sqrt{100}} = (0.610, 0.670).[/latex]
Interpretation: we can be 95% confident that the mean ABI of women with peripheral arterial disease is somewhere between 0.610 and 0.670. The entire interval is below 0.9, so we can claim [latex]\mu < 0.9.[/latex]  Yes, since a healthy (or normal) ABI is 0.9 or greater; however, 0.9 is outside the confidence interval (it is above the entire interval), so we can claim that the mean ABI of women with peripheral arterial disease is below 0.9, i.e., [latex]\mu < 0.9[/latex]. This is consistent with the conclusion of the ttest in part a).
 Assumptions: