Question

g A random sample of size 16 taken from a normally distributed population revealed a sample mean of 50 and a sample variance of 36. The upper limit of a 95% confidence interval for the population mean would equal:

Answer 1

The upper limit is

`k = 52.94`

Explanation:

From the question we told that

The sample size is
`n = 16`

The sample mean is
`\= x = 50`

The sample variance is
`\sigma ^2 = 36`

For a 95% confidence interval the confidence level is 95%

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

`\alpha = 100 - 95`

`\alpha = 5 \%`

`\alpha = 0.05`

Next we obtain the critical value of
`(\alpha )/(2)` from the normal distribution table(reference- math dot armstrong dot edu), the value is

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

Generally the margin of error is mathematically represented as

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

Here
`\sigma` is the standard deviation which is mathematically evaluated as

`\sigma = √(\sigma^2)`

substituting values

`\sigma = √(36)`

=>
`\sigma = 6`

So

`E = 1.96 * (6)/(√(16) )`

`E = 2.94`

The 95% confidence interval is mathematically represented as

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

substituting values

`50 -2.94 < \mu <50 +2.94`

`47.06 < \mu <52.94`

The upper limit is

`k = 52.94`