Question

Two hundred undergraduate students were randomly selected from a university that has 47,000 students in total. Systolic blood pressure was tested on the 200 students. The sample mean is 118.0 mmHg and the sample standard deviation is 11.0 mmHg. Please construct a 95% confidence interval for the population mean of students' systolic blood pressure. Which one of the following results is the closest to your answer? (Hint: use 1.96 as the critical z-value)

a. [110.5, 125.5]
b. [112.5, 123.5]
c. [114.5, 121.5]
d. [116.5, 119.5]

Answer 1

The correct option is d

Explanation:

From the question we are told that

The population size is
`N = 47000`

The sample size is
`n = 200`

The sample mean is
`\= x = 118.0 \ mmHg`

The standard deviation is
`\sigma = 11.0 \ mmHg`

Given that the confidence level is 95% then the level of significance can be calculated as

`\alpha = 100 - 95`

`\alpha = 5 \%`

`\alpha = 0.05`

Next we obtain the critical value of
`(\alpha )/(2)` from z-table , the value is
`Z_{(\alpha )/(2) } = Z_{(0.05 )/(2) } = 1.96`

The reason we are obtaining critical value of
`(\alpha )/(2)` instead of
`\alpha` is because

`\alpha` represents the area under the normal curve where the confidence level interval (
`1-\alpha`) did not cover which include both the left and right tail while

`(\alpha )/(2)` is just the area of one tail which what we required to calculate the margin of error .

NOTE: We can also obtain the value using critical value calculator (math dot armstrong dot edu)

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *(\sigma )/( √(n) )`

substituting values

`E = 1.96 *(11.0 )/( √(200) )`

`E = 1.5245`

The 95% confidence level interval is mathematically represented as

`\= x - E < \mu < \= x + E`

substituting values

`118.0 - 1.5245 < \mu < 118.0 + 1.5245`

`116.5< \mu < 119.5`

`[116.5 , 119.5]`