Question
At a local Mcdonald’s managers will return a shipment of hamburger buns if more than 10% of the buns are crushed. A random sample of 82 buns, 15 buns crashed. Conduct a 1% significance test to determine if the shipment should be accepted or not. Make you decision.
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Answered by
GPT-4o mini
To assess whether the shipment of hamburger buns should be accepted or returned, we can conduct a hypothesis test for the proportion of crushed buns. Here are the steps we'll follow:
1. **Define the null and alternative hypotheses:**
- Null Hypothesis (\(H_0\)): \(p \leq 0.10\) (the proportion of crushed buns is less than or equal to 10%)
- Alternative Hypothesis (\(H_a\)): \(p > 0.10\) (the proportion of crushed buns is greater than 10%)
2. **Collect data:**
- Sample size (\(n\)) = 82
- Number of crushed buns (\(x\)) = 15
- Sample proportion (\(\hat{p}\)) = \(\frac{x}{n} = \frac{15}{82} \approx 0.1829\)
3. **Calculate the test statistic:**
The test statistic for a proportion can be calculated using the formula:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
\]
where \(p_0 = 0.10\) (the value under the null hypothesis).
Plugging in the values:
\[
z = \frac{0.1829 - 0.10}{\sqrt{\frac{0.10 \cdot (1 - 0.10)}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{\frac{0.10 \cdot 0.90}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{\frac{0.09}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{0.00109756}} \approx \frac{0.0829}{0.0331} \approx 2.50
\]
4. **Determine the critical value:**
For a one-tailed test at the 1% significance level (\(\alpha = 0.01\)), we look up the critical value from the standard normal distribution table. The critical z-value for \(\alpha = 0.01\) is approximately 2.33.
5. **Make a decision:**
- If the calculated \(z\) statistic is greater than the critical value, we reject the null hypothesis.
- Here, \(z \approx 2.50\) > 2.33.
Since the calculated \(z\) statistic exceeds the critical value, we reject the null hypothesis.
6. **Conclusion:**
Based on the results of the hypothesis test, we reject the null hypothesis that the proportion of crushed buns is less than or equal to 10%. Thus, we conclude that there is sufficient evidence to claim that more than 10% of the buns are crushed, and the shipment should be returned.
1. **Define the null and alternative hypotheses:**
- Null Hypothesis (\(H_0\)): \(p \leq 0.10\) (the proportion of crushed buns is less than or equal to 10%)
- Alternative Hypothesis (\(H_a\)): \(p > 0.10\) (the proportion of crushed buns is greater than 10%)
2. **Collect data:**
- Sample size (\(n\)) = 82
- Number of crushed buns (\(x\)) = 15
- Sample proportion (\(\hat{p}\)) = \(\frac{x}{n} = \frac{15}{82} \approx 0.1829\)
3. **Calculate the test statistic:**
The test statistic for a proportion can be calculated using the formula:
\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}
\]
where \(p_0 = 0.10\) (the value under the null hypothesis).
Plugging in the values:
\[
z = \frac{0.1829 - 0.10}{\sqrt{\frac{0.10 \cdot (1 - 0.10)}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{\frac{0.10 \cdot 0.90}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{\frac{0.09}{82}}}
\]
\[
z = \frac{0.0829}{\sqrt{0.00109756}} \approx \frac{0.0829}{0.0331} \approx 2.50
\]
4. **Determine the critical value:**
For a one-tailed test at the 1% significance level (\(\alpha = 0.01\)), we look up the critical value from the standard normal distribution table. The critical z-value for \(\alpha = 0.01\) is approximately 2.33.
5. **Make a decision:**
- If the calculated \(z\) statistic is greater than the critical value, we reject the null hypothesis.
- Here, \(z \approx 2.50\) > 2.33.
Since the calculated \(z\) statistic exceeds the critical value, we reject the null hypothesis.
6. **Conclusion:**
Based on the results of the hypothesis test, we reject the null hypothesis that the proportion of crushed buns is less than or equal to 10%. Thus, we conclude that there is sufficient evidence to claim that more than 10% of the buns are crushed, and the shipment should be returned.
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