Internet service: An Internet service provider sam

Questions: Internet service: An Internet service provider sampled 530 customers and found that 70 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 0.132. Part: 1 / 3 Part 2 of 3 (b) Construct a 90% confidence interval for the proportion of all customers who experienced an interruption. Round the answer to at least three decimal places. A 90% confidence interval for the proportion of all customers who experienced an interruption is <p<

Transcript text: Internet service: An Internet service provider sampled 530 customers and found that 70 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 0.132 . Part: $1 / 3$ Part 2 of 3 (b) Construct a $90 \%$ confidence interval for the proportion of all customers who experienced an interruption. Round the answer to at least three decimal places. A $90 \%$ confidence interval for the proportion of all customers who experienced an interruption is $\square$ $

Solution

Solution Steps

Step 1: Finding a Point Estimate for the Population Proportion

To find the point estimate for the population proportion, we use the formula $p = \frac{x}{n}$, where $x$ is the number of successes in the sample and $n$ is the sample size. In this case, $p = \frac{70}{530} = 0.132$.

Step 2: Constructing a Confidence Interval for the Population Proportion

The standard error of the proportion is calculated using the formula $SE = \sqrt{\frac{p(1-p)}{n}}$, giving $SE = \sqrt{\frac{0.132(0.868)}{530}} = 0.015$.

With a Z-score of 1.645, the margin of error is $ME = Z \times SE = 1.645 \times 0.015 = 0.024$.

Thus, the confidence interval is constructed as $p \pm ME = 0.132 \pm 0.024$, which gives the interval [0.108, 0.156].