nykid2000
Answered



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?

Verified Answer :

toporc

The 98% confidence interval is found from the formula:
\(C.I.=\bar{x}+-2.33\frac{\sigma}{ \sqrt{n} }\)
Plugging in the given values, we get:
\(C.I.=82+-\frac{5}{ \sqrt{235} }\)
The result is (81.2 - 82.8).

Answer:

Option C

Step-by-step 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 =\(\frac{5}{\sqrt235} } =0.3262\)

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

Related Questions