Question
An Internet service provider sampled 542 customers and found that 66 of them experienced an interruption in high-speed service during the previous month.
(a) Find a point estimate for the population of all customers who experienced an interruption. Round the answer to at least three decimal places.
(a) Find a point estimate for the population of all customers who experienced an interruption. Round the answer to at least three decimal places.
Answers
Answer
(b) Construct a 99% confidence interval for the proportion of all customers who experienced an interruption. Round the answer to at least three decimal places.
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Answer
(c) The company's quality control manager claims that no more than 7% of its customers experienced an interruption during the previous month. Does the confidence interval contradict this claim? Explain.
Answered by
GPT-4o mini
To find the point estimate for the proportion of customers who experienced an interruption in high-speed service, you can use the formula for the sample proportion \( \hat{p} \):
\[
\hat{p} = \frac{x}{n}
\]
where:
- \( x \) is the number of customers who experienced an interruption (66 in this case),
- \( n \) is the total number of customers sampled (542 in this case).
Substituting the values:
\[
\hat{p} = \frac{66}{542} \approx 0.121
\]
To round this to at least three decimal places, we get:
\[
\hat{p} \approx 0.121
\]
Thus, the point estimate for the proportion of all customers who experienced an interruption in service is approximately **0.121**.
\[
\hat{p} = \frac{x}{n}
\]
where:
- \( x \) is the number of customers who experienced an interruption (66 in this case),
- \( n \) is the total number of customers sampled (542 in this case).
Substituting the values:
\[
\hat{p} = \frac{66}{542} \approx 0.121
\]
To round this to at least three decimal places, we get:
\[
\hat{p} \approx 0.121
\]
Thus, the point estimate for the proportion of all customers who experienced an interruption in service is approximately **0.121**.
Answered by
GPT-4o mini
To construct a 99% confidence interval for the proportion of customers who experienced an interruption, we can follow these steps:
1. **Calculate the sample proportion (\( \hat{p} \)):**
We already calculated \( \hat{p} \) in the previous part:
\[
\hat{p} = \frac{66}{542} \approx 0.121
\]
2. **Calculate the standard error (SE) of the proportion:**
The formula for the standard error of the proportion is:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
where \( n \) is the sample size (542) and \( \hat{p} \) is the sample proportion (0.121).
Let's calculate that:
\[
SE = \sqrt{\frac{0.121(1 - 0.121)}{542}} \approx \sqrt{\frac{0.121 \times 0.879}{542}} \approx \sqrt{\frac{0.106239}{542}} \approx \sqrt{0.000196}}
\]
\[
SE \approx 0.014
\]
3. **Find the Z-score for a 99% confidence level:**
For a 99% confidence level, the Z-score (critical value) is approximately 2.576.
4. **Calculate the margin of error (ME):**
The margin of error can be calculated using:
\[
ME = Z \times SE
\]
Substituting the values:
\[
ME = 2.576 \times 0.014 \approx 0.036144
\]
5. **Construct the confidence interval:**
The confidence interval can be constructed using the formula:
\[
\hat{p} \pm ME
\]
Calculating the lower and upper limits:
- Lower limit: \( 0.121 - 0.036144 \approx 0.085856 \)
- Upper limit: \( 0.121 + 0.036144 \approx 0.157144 \)
Finally, rounding to at least three decimal places, we get:
\[
0.086 < p < 0.157
\]
Thus, the 99% confidence interval for the proportion of all customers who experienced an interruption is approximately **(0.086, 0.157)**.
1. **Calculate the sample proportion (\( \hat{p} \)):**
We already calculated \( \hat{p} \) in the previous part:
\[
\hat{p} = \frac{66}{542} \approx 0.121
\]
2. **Calculate the standard error (SE) of the proportion:**
The formula for the standard error of the proportion is:
\[
SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}
\]
where \( n \) is the sample size (542) and \( \hat{p} \) is the sample proportion (0.121).
Let's calculate that:
\[
SE = \sqrt{\frac{0.121(1 - 0.121)}{542}} \approx \sqrt{\frac{0.121 \times 0.879}{542}} \approx \sqrt{\frac{0.106239}{542}} \approx \sqrt{0.000196}}
\]
\[
SE \approx 0.014
\]
3. **Find the Z-score for a 99% confidence level:**
For a 99% confidence level, the Z-score (critical value) is approximately 2.576.
4. **Calculate the margin of error (ME):**
The margin of error can be calculated using:
\[
ME = Z \times SE
\]
Substituting the values:
\[
ME = 2.576 \times 0.014 \approx 0.036144
\]
5. **Construct the confidence interval:**
The confidence interval can be constructed using the formula:
\[
\hat{p} \pm ME
\]
Calculating the lower and upper limits:
- Lower limit: \( 0.121 - 0.036144 \approx 0.085856 \)
- Upper limit: \( 0.121 + 0.036144 \approx 0.157144 \)
Finally, rounding to at least three decimal places, we get:
\[
0.086 < p < 0.157
\]
Thus, the 99% confidence interval for the proportion of all customers who experienced an interruption is approximately **(0.086, 0.157)**.
Answered by
GPT-4o mini
To determine whether the company's quality control manager's claim that no more than 7% of its customers experienced an interruption contradicts the confidence interval we calculated, we need to compare the claim with the confidence interval.
### Summary of the Confidence Interval
The 99% confidence interval we constructed for the proportion of customers who experienced an interruption is approximately:
\[
0.086 < p < 0.157
\]
### Comparison with the Claim
The manager's claim is that no more than 7% (or 0.07) of customers experienced an interruption.
### Analysis:
- The **upper bound** of the confidence interval is approximately **0.157**, which is significantly higher than **0.07**.
- The **lower bound** of the confidence interval is approximately **0.086**, which is also above **0.07**.
### Conclusion:
Since the entire confidence interval (0.086 to 0.157) lies above the claim of 0.07, the confidence interval does indeed **contradict** the manager's claim.
### Explanation:
This contradiction indicates that we have strong statistical evidence suggesting that a greater proportion of customers (between 8.6% and 15.7%) experienced interruptions than the manager's assertion of no more than 7%. Therefore, the claim made by the quality control manager is not supported by the data from the sample.
### Summary of the Confidence Interval
The 99% confidence interval we constructed for the proportion of customers who experienced an interruption is approximately:
\[
0.086 < p < 0.157
\]
### Comparison with the Claim
The manager's claim is that no more than 7% (or 0.07) of customers experienced an interruption.
### Analysis:
- The **upper bound** of the confidence interval is approximately **0.157**, which is significantly higher than **0.07**.
- The **lower bound** of the confidence interval is approximately **0.086**, which is also above **0.07**.
### Conclusion:
Since the entire confidence interval (0.086 to 0.157) lies above the claim of 0.07, the confidence interval does indeed **contradict** the manager's claim.
### Explanation:
This contradiction indicates that we have strong statistical evidence suggesting that a greater proportion of customers (between 8.6% and 15.7%) experienced interruptions than the manager's assertion of no more than 7%. Therefore, the claim made by the quality control manager is not supported by the data from the sample.
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