Math  /  Data & Statistics

QuestionInternet service: An Internet service provider sampled 538 customers and found that 78 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 0.145 .
Part: 1/31 / 3
Part 2 of 3 (b) Construct an 80%80 \% confidence interval for the proportion of all customers who experienced an interruption. Round the answer to at least three decimal places.
An 80%80 \% confidence interval for the proportion of all customers who experienced an interruption is \square <p<<p< \square

Studdy Solution

STEP 1

What is this asking? Out of 538 internet customers, 78 had their internet cut out last month.
We need to estimate how likely it is that *any* customer had an outage, and give a range we're 80% sure the real answer lies within. Watch out! Don't mix up the number of customers surveyed with the number who had problems!
Also, remember that a confidence interval isn't a guarantee, it's a probabilistic statement.

STEP 2

1. Calculate the point estimate.
2. Find the margin of error.
3. Construct the confidence interval.

STEP 3

Let's **dive in**!
The *best guess* we have for the true proportion of customers with outages is just the proportion in our sample.
We'll call this our **point estimate**.

STEP 4

We have 7878 customers with outages out of a total of 538538 customers surveyed.
So, our **point estimate** is: 785380.145 \frac{78}{538} \approx \mathbf{0.145}

STEP 5

Now, let's find the **margin of error**.
This tells us how much our point estimate might be off by.
It depends on how confident we want to be (80% in this case) and the size of our sample.

STEP 6

For an 80% confidence level, the critical *z*-value is about 1.28 \mathbf{1.28} .
You can find this using a *z*-table or calculator.
This *z*-value tells us how many standard deviations wide our confidence interval needs to be to capture 80% of the possible outcomes.

STEP 7

The formula for the **margin of error** is: zp(1p)n z \cdot \sqrt{\frac{p(1-p)}{n}} where zz is the *z*-value, pp is our **point estimate**, and nn is the sample size.

STEP 8

Let's plug in our values: 1.280.145(10.145)538 1.28 \cdot \sqrt{\frac{0.145(1 - 0.145)}{538}} 1.280.1450.855538 1.28 \cdot \sqrt{\frac{0.145 \cdot 0.855}{538}} 1.280.123975538 1.28 \cdot \sqrt{\frac{0.123975}{538}} 1.280.0002304368 1.28 \cdot \sqrt{0.0002304368} 1.280.015180.0194 1.28 \cdot 0.01518 \approx \mathbf{0.0194} So, our **margin of error** is approximately 0.0194\mathbf{0.0194}.

STEP 9

Finally, let's build that **confidence interval**!
We take our **point estimate** and add/subtract the **margin of error** to get the lower and upper bounds.

STEP 10

**Lower Bound:** 0.1450.0194=0.1256 0.145 - 0.0194 = \mathbf{0.1256}

STEP 11

**Upper Bound:** 0.145+0.0194=0.1644 0.145 + 0.0194 = \mathbf{0.1644}

STEP 12

We are 80% confident that the true proportion of customers who experienced an interruption is between 0.126\mathbf{0.126} and 0.164\mathbf{0.164}.

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