Question

7.2.9-T Question Help Here are summary statistics for randomly selected weights of newborn girls: n= 157, x = 33.6 hg, s = 6.2 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 32.3 hg​

Answer 1

The 98% confidence interval estimate of the mean is between 32.45 hg and 34.75 hg.

32.3 is not a part of the confidence interval, which means that yes, these results are very different from the confidence interval 32.3 hg​

Explanation:

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

`\alpha = (1 - 0.98)/(2) = 0.01`

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 - 0.01 = 0.99`, so Z = 2.327.

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 = 2.327(6.2)/(√(157)) = 1.15`

The lower end of the interval is the sample mean subtracted by M. So it is 33.6 - 1.15 = 32.45 hg

The upper end of the interval is the sample mean added to M. So it is 33.6 + 1.15 = 34.75 hg

The 98% confidence interval estimate of the mean is between 32.45 hg and 34.75 hg.

Are these results very different from the confidence interval 32.3 hg​

32.3 is not a part of the confidence interval, which means that yes, these results are very different from the confidence interval 32.3 hg​