Question

salma recently switched her primary doctor to one specializing in caring for elderly patients. on her new doctors website, it says that the mean systolic blood pressure among elderly females is 115 millimeters of mercury (mmhg). salma believes the value is actually higher. she bases her belief on a recently reported study of 18 randomly selected, elderly females. the sample mean systolic blood pressure was 120 mmhg, and the sample standard deviation was 22 mmhg. assume that the systolic blood pressures of elderly females are approximately normally distributed. based on the study, at the 0.10 level of significance, can it be concluded that μ, the population mean systolic blood pressure among elderly females, is greater than 115 mmhg? perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) can it be concluded that the mean systolic blood pressure among elderly females is greater than 115 mmhg? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is the statement of no - effect. The alternative hypothesis $H_1$ is what we are trying to find evidence for. Since we want to test if the population mean is greater than 115, $H_0:\mu = 115$ and $H_1:\mu>115$.

Step2: Determine test statistic type

We do not know the population standard deviation $\sigma$, and the sample size $n = 18$ is small ($n<30$). We assume the population is approximately normally distributed. So we use a t - test statistic, $t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$.

Step3: Calculate the test statistic

We are given $\bar{x}=120$, $\mu_0 = 115$, $s = 22$, and $n = 18$.
\[t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}=\frac{120 - 115}{22/\sqrt{18}}\approx\frac{5}{22/4.243}\approx\frac{5}{5.185}\approx0.964\]

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=18 - 1 = 17$. For a one - tailed t - test with $t = 0.964$ and $df=17$, using a t - distribution table or a calculator, the p - value is $P(T>0.964)\approx0.174$.

Step5: Make a decision

The level of significance is $\alpha=0.10$. Since the p - value ($0.174$) is greater than $\alpha = 0.10$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu = 115$; $H_1:\mu>115$
(b) t - test statistic
(c) $0.964$
(d) $0.174$
(e) No