Internet service: An Internet service provider sampled 530 customers and found that 80 of them experienced an interruption in high-speed service during the previous month.
Part 1 of 3
(a) Find a point estimate for the population of all customers who experienced an interruption. Round the answer to at least three decimal places.
The point estimate for the population proportion of all customers who experienced an interruption is 0.151 .
Part: $1 / 3$
Part 2 of 3
(b) Construct a $99.5 \%$ confidence interval for the proportion of all customers who experienced an interruption. Round the answer to at least three decimal places.
A $99.5 \%$ confidence interval for the proportion of all customers who experienced an interruption is $\square
Solution
Step 1 ：Given that an Internet service provider sampled 530 customers and found that 80 of them experienced an interruption in high-speed service during the previous month.
Step 2 ：Part 1: Find a point estimate for the population of all customers who experienced an interruption. The point estimate for the population proportion of all customers who experienced an interruption is \(0.151\).
Step 3 ：Part 2: Construct a \(99.5\%\) confidence interval for the proportion of all customers who experienced an interruption. The confidence interval can be calculated using the formula for the confidence interval of a proportion, which is \(p̂ ± Z*√((p̂(1-p̂))/n)\), where \(p̂\) is the sample proportion, \(n\) is the sample size, and \(Z\) is the Z-score corresponding to the desired level of confidence. In this case, \(p̂\) is \(0.151\) (80/530), \(n\) is 530, and \(Z\) is approximately \(2.807\) (for a \(99.5\%\) confidence interval).
Step 4 ：Using the above values, we calculate the standard error (se) as \(0.015550252522493096\).
Step 5 ：Then, we calculate the lower and upper bounds of the confidence interval as \(0.10729331228950345\) and \(0.19459348016332673\) respectively.
Step 6 ：Final Answer: A \(99.5\%\) confidence interval for the proportion of all customers who experienced an interruption is \(\boxed{0.107}
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Solution
Step 1 ：Given that an Internet service provider sampled 530 customers and found that 80 of them experienced an interruption in high-speed service during the previous month.
Step 2 ：Part 1: Find a point estimate for the population of all customers who experienced an interruption. The point estimate for the population proportion of all customers who experienced an interruption is \(0.151\).
Step 3 ：Part 2: Construct a \(99.5\%\) confidence interval for the proportion of all customers who experienced an interruption. The confidence interval can be calculated using the formula for the confidence interval of a proportion, which is \(p̂ ± Z*√((p̂(1-p̂))/n)\), where \(p̂\) is the sample proportion, \(n\) is the sample size, and \(Z\) is the Z-score corresponding to the desired level of confidence. In this case, \(p̂\) is \(0.151\) (80/530), \(n\) is 530, and \(Z\) is approximately \(2.807\) (for a \(99.5\%\) confidence interval).
Step 4 ：Using the above values, we calculate the standard error (se) as \(0.015550252522493096\).
Step 5 ：Then, we calculate the lower and upper bounds of the confidence interval as \(0.10729331228950345\) and \(0.19459348016332673\) respectively.
Step 6 ：Final Answer: A \(99.5\%\) confidence interval for the proportion of all customers who experienced an interruption is \(\boxed{0.107}
