Question

Recent incidents of food contamination have caused great concern among consumers. The article "How Safe Is That Chicken?" (Consumer Reports, Jan. 2010: 19-23) reported that 35 of 80 randomly selected Perdue brand broilers tested positively for either campylobacter or salmonella (or' both), the leading bacterial causes of food-borne disease, whereas 66 of 80 Tyson brand broilers tested positive.

1. Does it appear that the true proportion of noncontaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level .01.
2. If the true proportions of non-contaminated chickens for the Perdue and Tyson brands are .50 and .25, respectively, how likely is it that the null hypothesis of equal proportions will be rejected when a .01 significance level is used and the sample sizes are both 80?

Short answer

(a) There is sufficient evidence to support the claim that the true proportion of non-contaminated Perdue broilers differs from that for the Tyson brand.

(b) \(P(z < - 6.04\) or \(z > - 0.72) = 0.7642 = 76.42\% \)

Step-by-step solution

To carry out a test of hypotheses using a significance level

Given:

\({x_1} = 35\)

\(\begin{array}{l}{n_1} = 80\\{x_2} = 66\\{n_2} = 80\\\alpha = 0.01\end{array}\)

1. Claim: \({p_1} \ne {p_2}\)
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:

\({\hat p_1} = \frac{{{x_1}}}{{{n_1}}} = \frac{{35}}{{80}} \approx 0.4375\)

\(\begin{array}{l}{{\hat p}_2} = \frac{{{x_2}}}{{{n_2}}} = \frac{{66}}{{80}} \approx 0.825\\{{\hat p}_p} = \frac{{{x_1} + {x_2}}}{{{n_1} + {n_2}}} = \frac{{35 + 66}}{{80 + 80}} \approx 0.63125\end{array}\)

The critical values are the values corresponding to a probability of \(0.005/0.995\) in table A.3:

\(z = \pm 2.575\)

The rejection region then contains all values below -2.575 and all values above 2.575.

Determine the value of the test statistic:

\(z = \frac{{{{\hat p}_1} - {{\hat p}_2}}}{{\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} }} = \frac{{0.4375 - 0.825}}{{\sqrt {0.63125(1 - 0.63125)} \sqrt {\frac{1}{{80}} + \frac{1}{{80}}} }} \approx - 5.08\)

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

\( - 5.08 < - 2.575 \Rightarrow {\rm{ Reject }}{H_0}\)

There is sufficient evidence to support the claim that the true proportion of non-contaminated Perdue broilers differs from that for the Tyson brand.

To Determine the probability of rejecting the null hypothesis

b)

Given:

\(\begin{array}{l}{p_1} = 0.50\\{p_2} = 0.25\end{array}\)

The critical values are the values corresponding to a probability of \(0.005/0.995\)in table A.3:

\(z = \pm 2.575\)

Determine the difference in proportions that correspond with these z-values (assuming null hypothesis \({p_1} = {p_2}\) is true):

\({\hat p_1} - {\hat p_1} = \left( {{p_1} - {p_2}} \right) + z\sqrt {{{\hat p}_p}\left( {1 - {{\hat p}_p}} \right)} \sqrt {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} = 0 \pm 2.575\sqrt {0.63125(1 - 0.63125)} \sqrt {\frac{1}{{80}}} + \frac{1}{{80}} \approx \pm 0.1964\)

Determine the z-score corresponding with these difference in proportions, assuming that the alternative hypothesis is true (Note: We use \({p_1}\) and \({p_2}\)instead of\({\hat p_p}\), because \({\hat p_p}\)is unknown since we do not know the sample proportions):

\(\begin{array}{l}z = \frac{{{{\hat p}_1} - {{\hat p}_2} - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{{p_1}\left( {1 - {p_1}} \right)}}{{{n_1}}} + \frac{{{p_2}\left( {1 - {p_2}} \right)}}{{{n_2}}}} }} = \frac{{0.1964 - (0.50 - 0.25)}}{{\sqrt {\frac{{0.50(1 - 0.50)}}{{80}} + \frac{{0.25(1 - 0.25)}}{{80}}} }} \approx - 0.72\\z = \frac{{{{\hat p}_1} - {{\hat p}_2} - \left( {{p_1} - {p_2}} \right)}}{{\sqrt {\frac{{{p_1}\left( {1 - {p_1}} \right)}}{{{n_1}}} + \frac{{{p_2}\left( {1 - {p_2}} \right)}}{{{n_2}}}} }} = \frac{{ - 0.1964 - (0.50 - 0.25)}}{{\sqrt {\frac{{0.50(1 - 0.50)}}{{80}} + \frac{{0.25(1 - 0.25)}}{{80}}} }} \approx - 6.04\end{array}\)

Determine the probability of rejecting the null hypothesis using table A.3:

\(\begin{array}{l}P(z < - 6.04{\rm{ or }}z > - 0.72) = P(z < - 6.04) + P(z > - 0.72)\\ = P(z < - 6.04) + 1 - P(z < - 0.72) \approx 0 + 1 - 0.2358 = 0.7642 = 76.42\% \end{array}\)

Thus we have a 76.42 chance of rejecting the null hypothesis.

Final proof

(a) There is sufficient evidence to support the claim that the true proportion of non-contaminated Perdue broilers differs from that for the Tyson brand.

(b) \(P(z < - 6.04\)or \(z > - 0.72) = 0.7642 = 76.42\% \)