Question

You are given the sample mean mean and the population standard deviation. Use this information to construct 90% and 95%confidence intervals for the population mean. From a random sample of 57 dates , the mean record high daily temperature i. a certain city has a mean of 85.62 F assume the population standard deviation is 13.59 F​

Answer 1

a) The 90% confidence interval is approximately (82.47, 88.77) F.

b) The 95% confidence interval is approximately (82.28, 88.96) F.

A confidence interval refers to the probability that a population parameter will fall between two set values. It is the mean of an estimate plus and minus the variation in that estimate or the range of values, given a level of confidence.

To construct a confidence interval for the population mean, apply the formula as follows:

CI =
`[ \bar{x} \pm Z \left( (\sigma)/(√(n)) \right) ]`

Where:

CI = Confidence Interval

`(\bar{x})` = the sample mean

`(\sigma)` = the population standard deviation

`(n)` = the sample size

`(Z)` = the z-score corresponding to the desired confidence level

The z-scores are:

For a 90% confidence interval = 1.645

For a 95% confidence interval = 1.96.

Using the given values:

Sample mean
`(\bar{x})` = 85.62 F

Population standard deviation
`(\sigma)` = 13.59 F

Sample size
`(n)` = 57

For a 90% confidence interval:
`[ 85.62 \pm 1.645 \left( (13.59)/(√(57)) \right) ]`

For a 95% confidence interval:
`[ 85.62 \pm 1.96 \left( (13.59)/(√(57)) \right) ]`

Calculating the confidence intervals:

For the 90% confidence interval:
`[ 85.62 \pm 1.645 \left( (13.59)/(√(57)) \right) = 85.62 \pm 3.15 ]`

Thus, the 90% confidence interval is approximately (82.47, 88.77) F.

For the 95% confidence interval:
`[ 85.62 \pm 1.96 \left( (13.59)/(√(57)) \right) = 85.62 \pm 3.34 ]`

Thus, the 95% confidence interval is approximately (82.28, 88.96) F.