Question

One wishes to investigate the hypothesis that exercise reduces systolic blood pressure. A 100 person sample from a particular sedentary population indicates that μ = 120 mm Hg with a standard deviation of 8.7 mm Hg. A sample of 33 members of the local running club have a mean systolic pressure of 113 mm Hg and a standard deviation of 9.2 mm Hg.

a. State Ha c. What do you conclude at alpha = .01?
b. State Ho

Answer 1

Answer on Question #52612 – Math – Statistics and Probability

One wishes to investigate the hypothesis that exercise reduces systolic blood pressure. A 100 person sample from a particular sedentary population indicates that $\mu = 120 \mathrm{~mmHg}$ with a standard deviation of $8.7 \mathrm{~mmHg}$. A sample of 33 members of the local running club has a mean systolic pressure of $113 \mathrm{~mmHg}$ and a standard deviation of $9.2 \mathrm{~mmHg}$.

a. State Ha

b. State Ho

c. What do you conclude at alpha = .01?

Solution

a. $H_{a}:\mu_{1} < \mu_{2}$

b. $H_0: \mu_1 \geq \mu_2$

c.

We assume that variances are unequal.

Test statistic is

$$t = \frac {\overline {{x _ {1}}} - \overline {{x _ {2}}}}{\sqrt {\frac {s _ {1} ^ {2}}{n _ {1}} + \frac {s _ {2} ^ {2}}{n _ {2}}}} = \frac {113 - 120}{\sqrt {\frac {9.2^ {2}}{33} + \frac {8.7^ {2}}{100}}} = -3.84.$$

Critical value for $\alpha = 0.01$ and $100 + 33 - 2 = 131$ degrees of freedom from t-table is

$$t _ {crit} = 3.37.$$

We reject the null hypothesis at alpha = 0.01 significance level because test statistic $t = -3.84 < -t_{crit}$.

We can conclude at alpha = 0.01 significance level that exercise reduces systolic blood pressure.

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