QuestionConstruct a confidence interval for garlic's effect on LDL cholesterol. Mean change: 3.2, SD: 18.6. Find point estimate and $95\%$ CI.

Studdy Solution

STEP 1

Assumptions1. The sample size is47. The sample mean (best point estimate of the population mean) is3.3. The sample standard deviation is18.64. We are constructing a95% confidence interval

STEP 2

The best point estimate of the mean is the sample mean.
$\mu_{estimate} = \bar{x}$

STEP 3

Plug in the given value for the sample mean to find the best point estimate of the mean.
$\mu_{estimate} =3.2$ So, the best point estimate of the mean is3.2.

STEP 4

Next, we need to find the standard error of the mean. This is the standard deviation divided by the square root of the sample size.
$SE = \frac{s}{\sqrt{n}}$

STEP 5

Plug in the given values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{18.}{\sqrt{47}}$

STEP 6

Calculate the standard error.
$SE \approx2.71$

STEP 7

For a95% confidence interval, the critical value (Z) from the standard normal distribution is approximately1.96.

STEP 8

Now, we can calculate the margin of error for the confidence interval. This is the critical value times the standard error.
$ME = Z \times SE$

STEP 9

Plug in the values for the critical value and the standard error to calculate the margin of error.
$ME =.96 \times2.71$

STEP 10

Calculate the margin of error.
$ME \approx5.31$

STEP 11

Now, we can construct the confidence interval. This is the sample mean plus and minus the margin of error.
$CI = \bar{x} \pm ME$

STEP 12

Plug in the values for the sample mean and the margin of error to construct the confidence interval.
$CI =.2 \pm5.31$

STEP 13

Calculate the confidence interval.
$CI = (-2.11,8.51)$ The95% confidence interval estimate of the mean net change in LDL cholesterol after the Garlicin treatment is (-2.11,8.51). This means that we are95% confident that the true mean net change in LDL cholesterol after the Garlicin treatment is between -2.11 and8.51. The confidence interval includes zero, which suggests that the Garlicin treatment may not have a significant effect on lowering LDL cholesterol.

Was this helpful?

QuestionConstruct a confidence interval for garlic's effect on LDL cholesterol. Mean change: 3.2, SD: 18.6. Find point estimate and $95\%$ CI.

Studdy Solution

STEP 1

Assumptions1. The sample size is47. The sample mean (best point estimate of the population mean) is3.3. The sample standard deviation is18.64. We are constructing a95% confidence interval

STEP 2

The best point estimate of the mean is the sample mean.
$\mu_{estimate} = \bar{x}$

STEP 3

Plug in the given value for the sample mean to find the best point estimate of the mean.
$\mu_{estimate} =3.2$ So, the best point estimate of the mean is3.2.

STEP 4

Next, we need to find the standard error of the mean. This is the standard deviation divided by the square root of the sample size.
$SE = \frac{s}{\sqrt{n}}$

STEP 5

Plug in the given values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{18.}{\sqrt{47}}$

STEP 6

Calculate the standard error.
$SE \approx2.71$

STEP 7

For a95% confidence interval, the critical value (Z) from the standard normal distribution is approximately1.96.

STEP 8

Now, we can calculate the margin of error for the confidence interval. This is the critical value times the standard error.
$ME = Z \times SE$

STEP 9

Plug in the values for the critical value and the standard error to calculate the margin of error.
$ME =.96 \times2.71$

STEP 10

Calculate the margin of error.
$ME \approx5.31$

STEP 11

Now, we can construct the confidence interval. This is the sample mean plus and minus the margin of error.
$CI = \bar{x} \pm ME$

STEP 12

Plug in the values for the sample mean and the margin of error to construct the confidence interval.
$CI =.2 \pm5.31$

STEP 13

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionConstruct a confidence interval for garlic's effect on LDL cholesterol. Mean change: 3.2, SD: 18.6. Find point estimate and 95%95\%95% CI.

Studdy Solution

STEP 1

Assumptions1. The sample size is47. The sample mean (best point estimate of the population mean) is3.3. The sample standard deviation is18.64. We are constructing a95% confidence interval

STEP 2

The best point estimate of the mean is the sample mean.μestimate=xˉ\mu_{estimate} = \bar{x}μestimate​=xˉ

STEP 3

Plug in the given value for the sample mean to find the best point estimate of the mean.μestimate=3.2\mu_{estimate} =3.2μestimate​=3.2So, the best point estimate of the mean is3.2.

STEP 4

Next, we need to find the standard error of the mean. This is the standard deviation divided by the square root of the sample size.SE=snSE = \frac{s}{\sqrt{n}}SE=n​s​

STEP 5

Plug in the given values for the standard deviation and the sample size to calculate the standard error.SE=18.47SE = \frac{18.}{\sqrt{47}}SE=47​18.​

STEP 6

Calculate the standard error.SE≈2.71SE \approx2.71SE≈2.71

STEP 7

For a95% confidence interval, the critical value (Z) from the standard normal distribution is approximately1.96.

STEP 8

Now, we can calculate the margin of error for the confidence interval. This is the critical value times the standard error.ME=Z×SEME = Z \times SEME=Z×SE

STEP 9

Plug in the values for the critical value and the standard error to calculate the margin of error.ME=.96×2.71ME =.96 \times2.71ME=.96×2.71

STEP 10

Calculate the margin of error.ME≈5.31ME \approx5.31ME≈5.31

STEP 11

Now, we can construct the confidence interval. This is the sample mean plus and minus the margin of error.CI=xˉ±MECI = \bar{x} \pm MECI=xˉ±ME

STEP 12

Plug in the values for the sample mean and the margin of error to construct the confidence interval.CI=.2±5.31CI =.2 \pm5.31CI=.2±5.31

STEP 13

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionConstruct a confidence interval for garlic's effect on LDL cholesterol. Mean change: 3.2, SD: 18.6. Find point estimate and $95\%$ CI.

The best point estimate of the mean is the sample mean.
$\mu_{estimate} = \bar{x}$

Plug in the given value for the sample mean to find the best point estimate of the mean.
$\mu_{estimate} =3.2$ So, the best point estimate of the mean is3.2.

Next, we need to find the standard error of the mean. This is the standard deviation divided by the square root of the sample size.
$SE = \frac{s}{\sqrt{n}}$

Plug in the given values for the standard deviation and the sample size to calculate the standard error.
$SE = \frac{18.}{\sqrt{47}}$

Calculate the standard error.
$SE \approx2.71$

Now, we can calculate the margin of error for the confidence interval. This is the critical value times the standard error.
$ME = Z \times SE$

Plug in the values for the critical value and the standard error to calculate the margin of error.
$ME =.96 \times2.71$

Calculate the margin of error.
$ME \approx5.31$

Now, we can construct the confidence interval. This is the sample mean plus and minus the margin of error.
$CI = \bar{x} \pm ME$

Plug in the values for the sample mean and the margin of error to construct the confidence interval.
$CI =.2 \pm5.31$