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 small - sample (n < 30) confidence interval when the population standard - deviation is unknown, we use the t - distribution. The formula for the confidence interval is $\bar{x}\pm t_{\alpha/2}\frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the critical value, $s$ is the sample standard deviation, and $n$ is the sample size.

Step2: Determine the degrees of freedom and $\alpha$

The sample size $n = 15$, so the degrees of freedom $df=n - 1=15 - 1 = 14$. The confidence level is 90%, so $\alpha=1 - 0.90 = 0.10$ and $\alpha/2=0.05$.

Step3: Find the critical value $t_{\alpha/2}$

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

Step4: Calculate the margin of error

The sample mean $\bar{x}=69$, the sample standard deviation $s = 2.5$, and $n = 15$. The margin of error $E=t_{\alpha/2}\frac{s}{\sqrt{n}}=1.761\times\frac{2.5}{\sqrt{15}}\approx1.144$.

Step5: Calculate the lower and upper limits

The lower limit is $\bar{x}-E=69 - 1.144\approx67.9$.
The upper limit is $\bar{x}+E=69 + 1.144\approx70.1$.

Answer:

Lower limit: 67.9
Upper limit: 70.1