Question

jane 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 120 millimeters of mercury (mmhg). jane believes the value is actually higher. she bases her belief on a recently reported study of 20 randomly selected, elderly females. the sample mean systolic blood pressure was 133 mmhg, and the sample standard deviation was 25 mmhg. assume that the systolic blood pressures of elderly females are approximately normally distributed. based on the study, at the 0.05 level of significance, can it be concluded that μ, the population mean systolic blood pressure among elderly females, is greater than 120 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 120 mmhg? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the population mean $\mu$ is equal to 120, and the alternative hypothesis $H_1$ is that the population mean $\mu$ is greater than 120. So, $H_0:\mu = 120$, $H_1:\mu>120$.

Step2: Determine test - statistic type

Since the population standard deviation $\sigma$ is unknown and we have a sample standard deviation $s$, and the population is approximately normally distributed, we use a t - test statistic. The formula for the t - test statistic is $t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized population mean, $s$ is the sample standard deviation, and $n$ is the sample size.

Step3: Calculate the test - statistic value

We are given that $\bar{x} = 133$, $\mu_0=120$, $s = 25$, and $n = 20$.
\[

$$\begin{align*} t&=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}\\ &=\frac{133 - 120}{25/\sqrt{20}}\\ &=\frac{13}{25/\sqrt{20}}\\ &=\frac{13}{25/4.472}\\ &=\frac{13}{5.590}\\ &\approx2.326 \end{align*}$$

\]

Step4: Calculate the p - value

The degrees of freedom is $df=n - 1=20 - 1 = 19$. Since this is a one - tailed test with $t\approx2.326$ and $df = 19$, using a t - distribution table or a calculator, the p - value is $P(t_{19}>2.326)\approx0.016$.

Step5: Make a decision

The level of significance is $\alpha = 0.05$. Since the p - value ($0.016$) is less than $\alpha(0.05)$, we reject the null hypothesis.

Answer:

(a) $H_0:\mu = 120$, $H_1:\mu>120$
(b) t - statistic
(c) $2.326$
(d) $0.016$
(e) Yes