Math / Data & Statistics

QuestioniClicker Question Systolic blood pressures of healthy adults follow a normal distribution. We would like to conduct a hypothesis test at the $5 \%$ level of significance to determine if the true mean systolic blood pressure of healthy adults is greater than 120 . We take a random sample of 25 healthy adults. The sample mean systolic blood pressure is calculated to be 122 and the sample standard deviation is calculated to be 7.8. The P -value and conclusion of the appropriate test of significance are, respectively: (A) between 0.05 and 0.10 , fail to reject $\mathrm{H}_{0}$ . (B) between 0.10 and 0.15 , reject $\mathrm{H}_{0}$ . (C) between 0.10 and 0.15 , fail to reject $\mathrm{H}_{0}$ . (D) between 0.15 and 0.20 , reject $\mathrm{H}_{0}$ . (E) between 0.15 and 0.20 , fail to $\mathrm{H}_{0}$ .

Studdy Solution

STEP 1

1. The population of systolic blood pressures follows a normal distribution.
2. We are conducting a one-sample t-test.
3. The null hypothesis $H_0$ is that the mean systolic blood pressure is 120.
4. The alternative hypothesis $H_a$ is that the mean systolic blood pressure is greater than 120.
5. The level of significance is $\alpha = 0.05$ .

STEP 2

1. State the hypotheses.
2. Calculate the test statistic.
3. Determine the P-value.
4. Compare the P-value to the significance level and make a conclusion.

STEP 3

State the null and alternative hypotheses:
\[ H_0: \mu = 120 $
\[ H_a: \mu > 120 $

STEP 4

Calculate the test statistic using the formula for a one-sample t-test:
$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$
where $\bar{x} = 122$ , $\mu_0 = 120$ , $s = 7.8$ , and $n = 25$ .
$t = \frac{122 - 120}{7.8/\sqrt{25}} = \frac{2}{1.56} \approx 1.282$

STEP 5

Determine the P-value for the calculated t-statistic with $n-1 = 24$ degrees of freedom.
Using a t-distribution table or calculator, find the P-value for $t = 1.282$ .
The P-value is between 0.10 and 0.15.

STEP 6

Compare the P-value to the significance level $\alpha = 0.05$ .
Since the P-value (between 0.10 and 0.15) is greater than 0.05, we fail to reject the null hypothesis $H_0$ .
The P-value and conclusion of the test are:
$\text{(C) between 0.10 and 0.15, fail to reject } H_0.$

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

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

Parents Influencer program Contact Policy Terms