Question

On day two of a study on body temperatures, 106 temperatures were taken. Suppose that we only have the first 10 temperatures to work with. The mean and standard deviation of these 10 scores were 98.44o F and 0.30o F, respectively. Construct a 95% confidence interval for the mean of all body temperatures.

Answer 1

Answer:

The 95% confidence interval = (98.225, 98.655)

Step-by-step explanation:

The formula for Confidence Interval =

Mean ± z × standard deviation/√n

Where Mean = 98.44°F

Standard deviation = 0.30°F

n = number of samples is 10

Because our number if samples is small, we use the t score confidence interval formula

Mean ± t × standard deviation/√n

Degrees of Freedom = n - 1

= 10 - 1 = 9

We find the t score of 95% confidence interval and degrees of freedom 9

= 2.262

Hence,

Confidence interval = 98.44 ± 2.262 × 0.30/√10

= 98.44 ± 0.214592162

Confidence Interval

98.44 - 0.214592162

= 98.225407838

≈ 98.225

98.44 + 0.214592162

= 98.654592162

≈ 98.655

The 95% confidence interval = (98.225, 98.655)