91Ó°ÊÓ

Given:

a)

\(\begin{array}{l}{{\hat p}_1} = 17.6\% = 0.176\\{n_1} = 529\\{{\hat p}_2} = 15.8\% = 0.158\\{n_2} = 563\\\alpha = 5\% = 0.05\end{array}\)

(a) Claim: \({p_1} \ne {p_2}\)

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states an equality. Since the null hypothesis is not the claim, the alternative hypothesis is the claim.

\(\begin{array}{l}{H_0}:{p_1} = {p_2}\\{H_a}:{p_1} \ne {p_2}\end{array}\)

The sample proportion is the number of successes divided by the sample size:

\(\begin{array}{l}{x_1} = {{\hat p}_1}{n_1} = 0.176(529) \approx 93\\{x_2} = {{\hat p}_2}{n_2} = 0.158(563) \approx 89\end{array}\)

\({\hat p_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}} = \frac{{93 + 89}}{{529 + 563}} = \frac{1}{6} \approx 0.1667\)

The critical values are the values corresponding to a probability of \(0.025/0.975\) in table A.3:

\(z = \pm 1.96\)

The rejection region then contains all values below -1.96 and all values above 1.96.

Determine the value of the test statistic:

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }} = \frac{{0.176 - 0.158}}{{\sqrt {0.1667(1 - 0.1667)} \sqrt {\frac{1}{{529}} + \frac{1}{{563}}} }} \approx 0.80\)

If the value of the test statistic is within the rejection region, then the null hypothesis is rejected:

\( - 1.96 < 0.80 < 1.96 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that the incidence rate of Gl problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group.

(b) Given: Sample sizes are equal.

\(\begin{array}{l}{p_1} = 15\% = 0.15\\{p_2} = 20\% = 0.20\\P\left( {{\rm{ Reject }}{H_0}\mid {H_0}{\rm{ false }}} \right) = 0.90\end{array}\)

\({q_1}\)is the complement of \({p_1}\)and \({q_2}\)is the complement of \({p_2}\).

\(\begin{array}{l}{q_1} = 1 - {p_1} = 1 - 0.15 = 0.85\\{q_2} = 1 - {p_2} = 1 - 0.20 = 0.80\end{array}\)

\(\beta \)is the probability of not rejecting the null hypothesis when the null hypothesis \({H_0}\)is false (use the complement rule):

\(\beta = 1 - P\left( {{\rm{ Reject }}{H_0}\mid {H_0}{\rm{ false }}} \right) = 1 - 0.90 = 0.10\)

The critical value of the significance level \(\alpha \) is the value corresponding to a probability of \(\alpha /2 = 0.025\) in table A.3:

\({z_{\alpha /2}} = 1.96\)

The critical value of the probability of a type II error \(\beta \) is the value corresponding to a probability of 0.10 in table A.3:

\({z_\beta } = 1.28\)

\(d\)is the difference in population means:

\(d = {p_1} - {p_2} = 0.15 - 0.20 = - 0.05\)

Formula sample size:

Fill in the known values and evaluate (round up!):\(n = {\frac{{1.96\sqrt {(0.15 + 0.20)(0.85 + 0.80)/2} + 1.28\sqrt {0.15(0.85) + 0.20(0.80)} }}{{{{( - 0.05)}^2}}}^2} \approx 1211\)\(\)\(\)

(a) There is not sufficient evidence to support the claim that the incidence rate of Gl problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control group.

(b) \(n = 1211\)\(\)