Question

In a study of 235 adults, the mean heart rate was 82 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 5 beats per minute. What is the 98% confidence interval for the mean beats per minute? 80.9-86.3, 70.9-83.3, 81.2-82.8, or 80.9-83.3?

Answer 1

The 98% confidence interval is found from the formula:

`C.I.=\bar{x}+-2.33(\sigma)/( √(n) )`
Plugging in the given values, we get:

`C.I.=82+-(5)/( √(235) )`
The result is (81.2 - 82.8).

Answer 2

Option C

Explanation:

Given that In a study of 235 adults, the mean heart rate was 82 beats per minute.

ie. sample mean= 82

Population std deviation = 5 beats per second

Since population std deviation is known, we can use z critical value for finding the confidence interval

Std dev = sigma = 5

Sample size n = 235

Std error of sample =
`(5)/(\sqrt235) } =0.3262`

z critical for 98%=2.33

Margin of error = 2.33 (std error) = 0.76

Hence confidence interval lower bound 82-0.76 = 81.24

and upper bound is 82+076 = 82.76

Round off to one decimal to get

(81.2, 82.8) i.e. option C is the right answer