Question

A similar follow-up study is done on a sample of 25 meat-lovers who never eat vegetables, again randomly selected from the same general population (population mean life expectancy = 75, population standard deviation = 5). This new sample of meat-eaters live to an average age of 77. What is the lower limit and upper limit of the 95% confidence interval for the life expectancy of this sample of meat-lovers?

Answer 1

The lower limit is 75.04

The upper limit is 78.96

Explanation:

From the question we are told that

The sample size is
`n = 25`

The sample mean is
`\= x = 77`

The standard deviation is
`\sigma = 5`

Given that the confidence level is 95% then the level of significance is mathematically represented as

`\alpha = (100 - 95)\%`

`\alpha = 0.05`

The critical value for
`(\alpha )/(2)` obtained from the normal distribution table is

`Z_{(\alpha )/(2) } = 1.96`

Generally the margin of error is mathematically represented as

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

=>
`E = 1.96* (5 )/(√(25) )`

=>
`E = 1.96`

The 95% confidence interval is mathematically represented as

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

=>
`77 - 1.96 < \mu < 77 + 1.96`

=>
`75.04 < \mu < 78.96`