Question

A random sample of 19 ​men's resting pulse rates showed a mean of 73.6 beats per minute and standard deviation of 18.2. Assume that pulse rates are Normally distributed. Use the table to complete parts​ (a) through​ (c) below. LOADING... Click the icon to view the​ t-table. a. Find a​ 95% confidence interval for the population mean pulse rate of​ men, and report it in a sentence. Choose the correct answer below​ and, if​ necessary, fill in the answer boxes to complete your choice. ​(Type integers or decimals rounded to the nearest tenth as needed. Use ascending​ order.)

A. We are​ 95% confident that the sample mean resting pulse rate is between nothing and ----------

B. We are​ 95% confident that the population mean resting pulse rate is between nothing and ---------

C. There is a​ 95% chance that all resting pulse rates will be between nothing and ------

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 73.6

sample standard deviation = s = 18.2

sample size = n = 19

Degrees of freedom = df = n - 1 = 19 - 1 = 18

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

t$\alpha$/2,df = t0.025,18 = 2.101

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.101 * (18.2 / $\sqrt$ 19)

Margin of error = E = 8.8

The 95% confidence interval estimate of the population mean is,

$\bar x$  ± E  

= 73.6  ± 8.8

= ( 64.8, 82.4 )

B. We are​ 95% confident that the population mean resting pulse rate is between 64.8 and 82.4.