Let's break down the scenario and answer questions 41–46 based on hypothesis testing principles:

Scenario Summary:

• The manufacturer claims a 7% defective rate.
• The client believes this rate is no longer accurate and tests the claim by examining a random sample of 632 items.
• In the sample, 62 items are found to be defective.

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Question 41: The parameter being investigated

The parameter being investigated is the true proportion of defective goods being sent currently to the client. We're interested in knowing if this rate has changed from the advertised 7%.

Answer: a) the fraction of damaged goods being sent currently to the client.

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Question 42: Null and alternative hypotheses

We are testing whether the defective rate has increased from the advertised 7%. The null hypothesis ($H_0$) is that the true proportion is still 7%, and the alternative hypothesis ($H_A$) is that the proportion has increased.

• $H_0: p = 0.07$
• $H_A: p > 0.07$ (one-tailed test, since the client suspects the defective rate is higher).

Answer: d) $H_0: p = 0.07$, $H_A: p > 0.07$

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Question 43: The z-statistic for this hypothesis test

The z-statistic is calculated using the formula: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$ where:

• $\hat{p} = \frac{62}{632} = 0.0981$ (sample proportion),
• $p_0 = 0.07$ (null hypothesis proportion),
• $n = 632$.

Calculating the z-statistic:

$z = \frac{0.0981 - 0.07}{\sqrt{\frac{0.07(1 - 0.07)}{632}}} = \frac{0.0281}{0.01014} \approx 2.77$

Answer: c) 2.77

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Question 44: P-value shading

Since this is a one-tailed test (with the alternative hypothesis $p > 0.07$), the p-value corresponds to the area to the right of the calculated z-value. Thus, the p-value picture would shade the area to the right of the z-statistic.

Answer: c) shades the area to the right, which is the area in a small tail.

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Question 45: The actual p-value

Using the z-value of 2.77, we find the p-value from standard normal tables or using a calculator. For $z = 2.77$, the area to the left is approximately 0.9972, so the area to the right (the p-value) is $1 - 0.9972 = 0.0028$.

Answer: c) 0.0028

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Question 46: Conclusion of the hypothesis test

The p-value (0.0028) is much smaller than the typical significance level ($\alpha = 0.05$). Therefore, we reject the null hypothesis. This means that we have sufficient evidence to conclude that the defective rate is significantly higher than the advertised 7%.

Answer: c) We have sufficient evidence of false advertising. The data in the sample suggest that the fraction of damaged goods recently being sent to the client has increased above the 7% stated rate.

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Follow-up questions:

1. How do you calculate the sample proportion from the data provided?
2. Why is a one-tailed test used in this scenario instead of a two-tailed test?
3. How does increasing the sample size affect the power of the hypothesis test?
4. What would happen if the p-value was larger than 0.05? What conclusion would we draw then?
5. What are the assumptions made in using a z-test for proportion?

Tip:

When performing a hypothesis test, always ensure that the sample size is large enough so that the sampling distribution of the sample proportion approximates a normal distribution (use the rule of thumb: $n \cdot p_0 \geq 10$ and $n \cdot (1 - p_0) \geq 10$).