An Internet service provider sampled 570 customers and found that 70 of them experienced an interruption in their service during the previous month. The company claims that no more than 10% Of customers experienced an interruption in the past month. Does the 95% confidence interval for The proportion support this claim? (Hint: you should first construct the 95% confidence interval for The proportion of customers who experienced an interruption.)
A) Yes
B) No
C) The reasonableness of the claim cannot be determined.

Question

Quizplus · Accepted Answer

To construct a 95% confidence interval for the proportion of customers who experienced an interruption, we first calculate the sample proportion $ p = \frac{70}{570} \approx 0.1228 $. The standard error (SE) for the proportion is calculated using the formula $ SE = \sqrt{\frac{p(1-p)}{n}} $, where $ n = 570 $. Plugging in the numbers, $ SE \approx \sqrt{\frac{0.1228(1-0.1228)}{570}} \approx 0.0146 $.For a 95% confidence interval, we use the Z-score for 95% confidence, which is approximately 1.96. The margin of error (ME) is $ ME = Z \times SE \approx 1.96 \times 0.0146 \approx 0.0286 $.The confidence interval is $ p \pm ME $, which is $ 0.1228 \pm 0.0286 $, resulting in an interval of approximately $ [0.0942, 0.1514] $.Since the interval includes proportions greater than 0.10, it does not support the company's claim that no more than 10% of customers experienced an interruption.

An Internet Service Provider Sampled 570 Customers and Found That