Question

A simple random sample of size 51 has sample mean of 70.72. The population distribution is approximately normal, with population standard deviation of 16.25. Determine the correct method of finding a 95% confidence interval for the unknown population mean and compute it. z-method: (66.26, 75.18) Cannot compute: the population size is too small. t-method: (66.15, 75.29) z-method: (66.15, 75.29)

Answer 1

z-method: (66.26, 75.18)

Explanation:

We have the standard deviation for the population, which means that we use the z-distribution to solve this question.

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(16.25)/(√(51)) = 4.46`

The lower end of the interval is the sample mean subtracted by M. So it is 70.72 - 4.46 = 66.26

The upper end of the interval is the sample mean added to M. So it is 70.72 + 4.46 = 75.18

The correct answer is z-method: (66.26, 75.18)