Question

suppose that a random sample of 15 adult u.s. males has a mean height of 69 inches with a standard deviation of 2.5 inches. if we assume that the heights of adult males in the u.s. are normally distributed, find a 90% confidence interval for the mean height of all u.s. males. give the lower limit and upper limit of the 90% confidence interval. carry your intermediate computations to at least three decimal places. round your answers to one decimal place. (if necessary, consult a list of formulas.) lower limit: upper limit:

Explanation:

Step1: Identify the formula

For a confidence interval when the population standard - deviation is unknown (we use the sample standard - deviation $s$ instead), the formula for a confidence interval of the population mean $\mu$ is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard - deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the critical value. Here, $\bar{x} = 69$, $s = 2.5$, $n = 15$. The degrees of freedom $df=n - 1=15 - 1 = 14$. For a 90% confidence interval, $\alpha=1 - 0.90 = 0.10$ and $\alpha/2=0.05$.

Step2: Find the critical value $t_{\alpha/2}$

Looking up the $t$ - value in the $t$ - distribution table with $df = 14$ and $\alpha/2 = 0.05$, we get $t_{0.05,14}=1.761$.

Step3: Calculate the margin of error $E$

The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}$. Substitute the values: $E = 1.761\times\frac{2.5}{\sqrt{15}}\approx1.761\times\frac{2.5}{3.873}\approx1.761\times0.646\approx1.138$.

Step4: Calculate the lower and upper limits

The lower limit is $\bar{x}-E=69 - 1.138\approx67.9$. The upper limit is $\bar{x}+E=69 + 1.138\approx70.1$.

Answer:

Lower limit: 67.9
Upper limit: 70.1