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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. a. Does it appear that the true proportion of non- contaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level.01 by obtaining a \(P\)-value. b. 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 ?

Step by step solution

01

Define Hypotheses

We will set up hypotheses for comparing the proportions of non-contaminated broilers between Perdue and Tyson. Let \( p_1 \) be the proportion of non-contaminated Perdue broilers and \( p_2 \) be the proportion for Tyson. The null hypothesis \( H_0 \) is that \( p_1 = p_2 \), and the alternative hypothesis \( H_a \) is that \( p_1 eq p_2 \).

02

Calculate Sample Proportions

For Perdue, 35 out of 80 broilers are contaminated, so \( 45 \) are not. Thus, the sample proportion for Perdue (\( \hat{p}_1 \)) is \( \frac{45}{80} = 0.5625 \). For Tyson, 66 out of 80 are contaminated, so \( 14 \) are not. Thus, \( \hat{p}_2 = \frac{14}{80} = 0.175 \).

03

Calculate Pooled Proportion

The pooled proportion \( \hat{p} \) is calculated using the formula \( \hat{p} = \frac{x_1 + x_2}{n_1 + n_2} \), where \( x_1 \) and \( x_2 \) are the numbers of non-contaminated broilers in each sample. So, \( \hat{p} = \frac{45 + 14}{80 + 80} = 0.36875 \).

04

Calculate Standard Error

The standard error (SE) for the difference in proportions is given by \( SE = \sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})} \), where \( n_1 \) and \( n_2 \) are the sample sizes. So, \( SE = \sqrt{0.36875 \times (1 - 0.36875) \times (\frac{1}{80} + \frac{1}{80})} = 0.0863 \).

05

Compute Test Statistic

The test statistic \( Z \) is \( \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.5625 - 0.175}{0.0863} \approx 4.49 \).

06

Find P-value

Using a Z-table or calculator, a \( Z \) value of \( 4.49 \) gives a \( P \)-value that is much smaller than the significance level of 0.01. This indicates strong evidence against the null hypothesis.

07

Conclusion for Part A

Since the \( P \)-value is less than 0.01, we reject the null hypothesis. There is significant evidence that the proportions of non-contaminated broilers differ between the two brands.

08

Calculate Power of Test for Part B

To calculate the power of the test, we need both the hypothesized proportions under the alternative, \( p_1 = 0.50 \) and \( p_2 = 0.25 \). Calculate the standard error for these proportions: \( SE_0 = \sqrt{0.375 \times 0.625 \times (\frac{1}{80} + \frac{1}{80})} \approx 0.088 \), and the standard error for the hypothesized difference: \( SE_A = \sqrt{0.5 \times 0.5 \times \frac{1}{80} + 0.25 \times 0.75 \times \frac{1}{80}} \approx 0.087 \).

09

Calculate Power via Test Statistic for Part B

Determine the Z value under the alternative hypothesis: \( Z = \frac{0.50 - 0.25}{0.087} \approx 2.87 \). From this Z value, using the significance level and standard power analysis, find \( 1 - \beta \), which shows the likelihood of rejecting the null when the alternative is true. This corresponds to significant success probabilities in standard normal distribution tables.

10

Conclusion for Part B

The likelihood to reject the null hypothesis, given \( p_1 = 0.50 \) and \( p_2 = 0.25 \), is quite high. Therefore, the power of the test is substantial.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proportion Test

A proportion test is a useful statistical tool for comparing two proportions to see if there's a significant difference between them. In our scenario with broilers, we want to know if the proportion of non-contaminated chickens of the Perdue brand is different from the Tyson brand. This involves calculating the sample proportion for each brand and comparing them. By setting the null hypothesis as both proportions being equal, the test aims to determine if any observed difference is due to random chance or a true difference in contamination rates.

Significance Level

The significance level, often denoted by \( \alpha \), is a critical component in hypothesis testing. It's essentially the threshold we set to decide whether to reject the null hypothesis. In this exercise, we use a significance level of 0.01, meaning we accept a 1% chance of incorrectly rejecting the null hypothesis when it's actually true. A smaller significance level, such as 0.01, provides stronger evidence against the null hypothesis if it is rejected, indicating a more reliable result.

P-value Calculation

The \( P \)-value helps us decide how likely it is that the observed data would occur if the null hypothesis were true. It's computed from the test statistic and compared against the significance level to make a decision. In the broiler contamination study, a very small \( P \)-value was calculated, indicating strong evidence against the null hypothesis. This suggests that the proportion of non-contaminated chicken differs significantly between the two brands.

Power of a Test

The power of a test is the probability that the test correctly rejects a false null hypothesis. It is fundamentally important because it tells you how likely you are to detect a difference when there is one. In hypothesis testing, it's vital to have high power to avoid Type II errors. In the exercise, the power calculation shows a high probability of correctly rejecting the null hypothesis if the true proportions are \( 0.50 \) for Perdue and \( 0.25 \) for Tyson. High power means the test is sensitive and reliable when the true difference exists.