Question

A medical researcher is studying the effects of a drug on blood pressure. Subjects in the study have their blood pressure taken at the beginning of the study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test. Change: Final Blood Pressure - Initial Blood Pressure The researcher wants to know if there is evidence that the drug increases blood pressure. At the end of 4 weeks, 32 subjects in the study had an average change in blood pressure of 2.5 with a standard deviation of 4.8. Find the p -value for the hypothesis test.

Answer 1

Answer:

The p -value for the hypothesis test is 0.003.

Step-by-step explanation:

In this case, a paired t-test would be used to determine whether the drug  increases blood pressure.

The hypothesis can be defined as follows:

H₀: The drug has no effect on blood pressure, i.e. d = 0.

Hₐ: The drug increases blood pressure, i.e. d > 0.

The information provided is:

$n=32\\\bar d=2.5\\S_{d}=4.8$

The test statistic is:

$t=\frac{\bar d}{S_{d}/\sqrt{n}}$

  $=\frac{2.5}{4.8/\sqrt{32}}\\\\=2.9463\\\\\approx 2.95$

The degrees of freedom of the test is:

$df=n-1=32-1=31$

The p-value of the test is:

$p-value=P(t_{31}>2.95)=0.003$

*Use a t-table.

Thus, the p -value for the hypothesis test is 0.003.

Answer 2

A t-test is an inferential statistic that is used to see if there is a significant difference in the means. The p-value for the hypothesis test is 0.003.

What is a t-test?

A t-test is an inferential statistic that is used to see if there is a significant difference in the means of two groups that are connected in some way.

As it is given that 32(n) subjects in the study had an average change in blood pressure of 2.5($\bar d$) with a standard deviation of 4.8($S_d$).

Now, in order to solve the problem we need to use t-test, the test will determine whether the blood pressure increases or not under the influence of the drug.

The hypothesis for the test can be defined as:

H₀: The drug has no effect on blood pressure, therefore, d = 0.

Hₐ: The drug has increased blood pressure, therefore, d>0.

The test static is:

$t = \dfrac{\bar d}{\dfrac{S_d}{\sqrt n}}\\\\\\t = \dfrac{2.5}{\dfrac{4.8}{32}} = 2.946 \approx 2.95$

Further, the degree of freedom of the test can be written as,

$d_f=n-1 = 32-1=31$

Thus, the p-value of the test is:

$p-value = P(t_{31} > 2.95)\\\\\text{Using the t-table}\\\\p-value = 0.003$

Hence, the p-value for the hypothesis test is 0.003.

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