An Internet service provider sampled 542 customers and found that 66 of them experienced an interruption in high-speed service during the previous month.

Question

(a) Find a point estimate for the population of all customers who experienced an interruption. Round the answer to at least three decimal places.

Answer 1

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.

Answer 2

(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.
[?] < p < [?]

Answer 3

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).

Answer 4

(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.

Answer 5

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.