Math / Data & Statistics

QuestionHomework \# 4: $7(1,2,3,4) 8(2,3,4)$ Question 21 of 40 (1 point) I Question Attempt: 1 of 3 Jonatha $\checkmark 12$ 13 $\checkmark 14$ 15 16 18 19 20 $\equiv 21$ $\checkmark 22$
Student loans: The Institute for College Access and Success reported that $65 \%$ of college students in a recent year graduated with student loan debt. A random sample of 90 graduates is drawn. Round your answers to at least four decimal places if necessary.
Part 1 of 6 (a) Find the mean $\mu_{\hat{p}}$ .
The mean $\mu_{p}$ is 0.65 .
Part: 1 / 6
Part 2 of 6 (b) Find the standard deviation $\sigma \hat{p}$ .
The standard deviation $\sigma \hat{p}$ is $\square$ Save For Later Subm

Studdy Solution

STEP 1

What is this asking? We're looking at a bunch of graduates and want to figure out, on average, what proportion of them have student loans, and how much that proportion varies from sample to sample. Watch out! Don't mix up the proportion of all students with loans and the proportion of a sample with loans.
Also, remember the difference between standard deviation and variance.

STEP 2

1. Find the mean proportion.
2. Calculate the standard deviation.

STEP 3

The problem tells us that 65% of college students graduate with student loan debt.
This is our population proportion, which we write as $p = 0.65$ .
When we take a random sample of graduates, we expect the sample proportion ( $\hat{p}$ ) to be the same as the population proportion, on average.

STEP 4

So, the mean of the sample proportion ( $\mu_{\hat{p}}$ ) is simply equal to the population proportion: $\mu_{\hat{p}} = p = \mathbf{0.65}$ This means that if we took tons of samples of graduates, the average of the proportions with loans in each sample would be around 0.65.

STEP 5

The standard deviation of the sample proportion ( $\sigma_{\hat{p}}$ ) tells us how spread out the sample proportions are likely to be.
The formula for this is: $\sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1-p)}{n}}$ where $p$ is the population proportion, and $n$ is the sample size.

STEP 6

We know that $p = 0.65$ and our sample size is $n = 90$ .
Let's plug these values into the formula: $\sigma_{\hat{p}} = \sqrt{\frac{0.65 \cdot (1-0.65)}{90}}$

STEP 7

First, let's calculate $1 - 0.65 = 0.35$ .
This represents the proportion of students without loans.
Then, we multiply this by $0.65$ to get $0.65 \cdot 0.35 = 0.2275$ .

STEP 8

Next, we divide this by our sample size of 90: $\frac{0.2275}{90} \approx 0.00252778$

STEP 9

Finally, we take the square root of this result: $\sqrt{0.00252778} \approx \mathbf{0.050277}$ So, the standard deviation of the sample proportion is approximately 0.0503 when rounded to four decimal places.

STEP 10

The mean of the sample proportion ( $\mu_{\hat{p}}$ ) is 0.65.
The standard deviation of the sample proportion ( $\sigma_{\hat{p}}$ ) is approximately 0.0503.

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Studdy Solution

STEP 1

STEP 2

1. Find the mean proportion.
2. Calculate the standard deviation.

STEP 3

The problem tells us that 65% of college students graduate with student loan debt.
This is our population proportion, which we write as $p = 0.65$ .
When we take a random sample of graduates, we expect the sample proportion ( $\hat{p}$ ) to be the same as the population proportion, on average.

STEP 4

So, the mean of the sample proportion ( $\mu_{\hat{p}}$ ) is simply equal to the population proportion: $\mu_{\hat{p}} = p = \mathbf{0.65}$ This means that if we took tons of samples of graduates, the average of the proportions with loans in each sample would be around 0.65.

STEP 5

The standard deviation of the sample proportion ( $\sigma_{\hat{p}}$ ) tells us how spread out the sample proportions are likely to be.
The formula for this is: $\sigma_{\hat{p}} = \sqrt{\frac{p \cdot (1-p)}{n}}$ where $p$ is the population proportion, and $n$ is the sample size.

STEP 6

We know that $p = 0.65$ and our sample size is $n = 90$ .
Let's plug these values into the formula: $\sigma_{\hat{p}} = \sqrt{\frac{0.65 \cdot (1-0.65)}{90}}$

STEP 7

First, let's calculate $1 - 0.65 = 0.35$ .
This represents the proportion of students without loans.
Then, we multiply this by $0.65$ to get $0.65 \cdot 0.35 = 0.2275$ .

STEP 8

Next, we divide this by our sample size of 90: $\frac{0.2275}{90} \approx 0.00252778$

STEP 9

Finally, we take the square root of this result: $\sqrt{0.00252778} \approx \mathbf{0.050277}$ So, the standard deviation of the sample proportion is approximately 0.0503 when rounded to four decimal places.

STEP 10

The mean of the sample proportion ( $\mu_{\hat{p}}$ ) is 0.65.
The standard deviation of the sample proportion ( $\sigma_{\hat{p}}$ ) is approximately 0.0503.

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Studdy Solution

STEP 1

STEP 2

1. Find the mean proportion.2. Calculate the standard deviation.

STEP 3

STEP 4

STEP 5

STEP 6

We know that p=0.65p = 0.65p=0.65 and our sample size is n=90n = 90n=90.Let's plug these values into the formula: σp^=0.65⋅(1−0.65)90 \sigma_{\hat{p}} = \sqrt{\frac{0.65 \cdot (1-0.65)}{90}} σp^​​=900.65⋅(1−0.65)​​

STEP 7

First, let's calculate 1−0.65=0.351 - 0.65 = 0.351−0.65=0.35.This represents the proportion of students *without* loans.Then, we multiply this by 0.650.650.65 to get 0.65⋅0.35=0.22750.65 \cdot 0.35 = 0.22750.65⋅0.35=0.2275.

STEP 8

Next, we divide this by our sample size of 90: 0.227590≈0.00252778 \frac{0.2275}{90} \approx 0.00252778 900.2275​≈0.00252778

STEP 9

Finally, we take the square root of this result: 0.00252778≈0.050277 \sqrt{0.00252778} \approx \mathbf{0.050277} 0.00252778​≈0.050277 So, the standard deviation of the sample proportion is approximately **0.0503** when rounded to four decimal places.

STEP 10

The mean of the sample proportion (μp^\mu_{\hat{p}}μp^​​) is **0.65**.The standard deviation of the sample proportion (σp^\sigma_{\hat{p}}σp^​​) is approximately **0.0503**.

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1. Find the mean proportion.
2. Calculate the standard deviation.

We know that $p = 0.65$ and our sample size is $n = 90$ .
Let's plug these values into the formula: $\sigma_{\hat{p}} = \sqrt{\frac{0.65 \cdot (1-0.65)}{90}}$

First, let's calculate $1 - 0.65 = 0.35$ .
This represents the proportion of students without loans.
Then, we multiply this by $0.65$ to get $0.65 \cdot 0.35 = 0.2275$ .

Next, we divide this by our sample size of 90: $\frac{0.2275}{90} \approx 0.00252778$

Finally, we take the square root of this result: $\sqrt{0.00252778} \approx \mathbf{0.050277}$ So, the standard deviation of the sample proportion is approximately 0.0503 when rounded to four decimal places.

The mean of the sample proportion ( $\mu_{\hat{p}}$ ) is 0.65.
The standard deviation of the sample proportion ( $\sigma_{\hat{p}}$ ) is approximately 0.0503.