Math / Data & Statistics

QuestionThe effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 64.4 for a sample of size 26 and standard deviation 19.4. Estimate how much the drug will lower a typical patient's systolic blood pressure (using a $95 \%$ confidence level). Assume the data is from a normally distributed population. Enter your answer as a tri-linear inequality accurate to three decimal places. $\square$ $<\mu<$ $\square$

Studdy Solution

STEP 1

1. The sample mean reduction in systolic blood pressure is $\bar{x} = 64.4$ .
2. The sample size is $n = 26$ .
3. The sample standard deviation is $s = 19.4$ .
4. The data is from a normally distributed population.
5. We are using a $95\%$ confidence level.

STEP 2

1. Calculate the standard error of the mean.
2. Determine the critical t-value for a $95\%$ confidence interval.
3. Calculate the margin of error.
4. Construct the confidence interval for the population mean.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{s}{\sqrt{n}}$
$\text{SEM} = \frac{19.4}{\sqrt{26}}$
$\text{SEM} \approx 3.804$

STEP 4

Determine the critical t-value for a $95\%$ confidence interval with $n-1 = 25$ degrees of freedom. Using a t-distribution table or calculator, find:
$t^* \approx 2.060$

STEP 5

Calculate the margin of error (ME) using the formula:
$\text{ME} = t^* \times \text{SEM}$
$\text{ME} = 2.060 \times 3.804$
$\text{ME} \approx 7.835$

STEP 6

Construct the confidence interval for the population mean $\mu$ :
$\bar{x} - \text{ME} < \mu < \bar{x} + \text{ME}$
$64.4 - 7.835 < \mu < 64.4 + 7.835$
$56.565 < \mu < 72.235$
The $95\%$ confidence interval for the reduction in systolic blood pressure is:
$\boxed{56.565 < \mu < 72.235}$

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Studdy Solution

STEP 1

1. The sample mean reduction in systolic blood pressure is $\bar{x} = 64.4$ .
2. The sample size is $n = 26$ .
3. The sample standard deviation is $s = 19.4$ .
4. The data is from a normally distributed population.
5. We are using a $95\%$ confidence level.

STEP 2

1. Calculate the standard error of the mean.
2. Determine the critical t-value for a $95\%$ confidence interval.
3. Calculate the margin of error.
4. Construct the confidence interval for the population mean.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{s}{\sqrt{n}}$
$\text{SEM} = \frac{19.4}{\sqrt{26}}$
$\text{SEM} \approx 3.804$

STEP 4

Determine the critical t-value for a $95\%$ confidence interval with $n-1 = 25$ degrees of freedom. Using a t-distribution table or calculator, find:
$t^* \approx 2.060$

STEP 5

Calculate the margin of error (ME) using the formula:
$\text{ME} = t^* \times \text{SEM}$
$\text{ME} = 2.060 \times 3.804$
$\text{ME} \approx 7.835$

STEP 6

Construct the confidence interval for the population mean $\mu$ :
$\bar{x} - \text{ME} < \mu < \bar{x} + \text{ME}$
$64.4 - 7.835 < \mu < 64.4 + 7.835$
$56.565 < \mu < 72.235$
The $95\%$ confidence interval for the reduction in systolic blood pressure is:
$\boxed{56.565 < \mu < 72.235}$

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Studdy Solution

STEP 1

1. The sample mean reduction in systolic blood pressure is xˉ=64.4 \bar{x} = 64.4 xˉ=64.4.2. The sample size is n=26 n = 26 n=26.3. The sample standard deviation is s=19.4 s = 19.4 s=19.4.4. The data is from a normally distributed population.5. We are using a 95% 95\% 95% confidence level.

STEP 2

1. Calculate the standard error of the mean.2. Determine the critical t-value for a 95% 95\% 95% confidence interval.3. Calculate the margin of error.4. Construct the confidence interval for the population mean.

STEP 3

Calculate the standard error of the mean (SEM) using the formula:SEM=sn \text{SEM} = \frac{s}{\sqrt{n}} SEM=n​s​SEM=19.426 \text{SEM} = \frac{19.4}{\sqrt{26}} SEM=26​19.4​SEM≈3.804 \text{SEM} \approx 3.804 SEM≈3.804

STEP 4

Determine the critical t-value for a 95% 95\% 95% confidence interval with n−1=25 n-1 = 25 n−1=25 degrees of freedom. Using a t-distribution table or calculator, find:t∗≈2.060 t^* \approx 2.060 t∗≈2.060

STEP 5

Calculate the margin of error (ME) using the formula:ME=t∗×SEM \text{ME} = t^* \times \text{SEM} ME=t∗×SEMME=2.060×3.804 \text{ME} = 2.060 \times 3.804 ME=2.060×3.804ME≈7.835 \text{ME} \approx 7.835 ME≈7.835

STEP 6

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1. The sample mean reduction in systolic blood pressure is $\bar{x} = 64.4$ .
2. The sample size is $n = 26$ .
3. The sample standard deviation is $s = 19.4$ .
4. The data is from a normally distributed population.
5. We are using a $95\%$ confidence level.

1. Calculate the standard error of the mean.
2. Determine the critical t-value for a $95\%$ confidence interval.
3. Calculate the margin of error.
4. Construct the confidence interval for the population mean.

Calculate the standard error of the mean (SEM) using the formula:
$\text{SEM} = \frac{s}{\sqrt{n}}$
$\text{SEM} = \frac{19.4}{\sqrt{26}}$
$\text{SEM} \approx 3.804$

Determine the critical t-value for a $95\%$ confidence interval with $n-1 = 25$ degrees of freedom. Using a t-distribution table or calculator, find:
$t^* \approx 2.060$

Calculate the margin of error (ME) using the formula:
$\text{ME} = t^* \times \text{SEM}$
$\text{ME} = 2.060 \times 3.804$
$\text{ME} \approx 7.835$