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​

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​

Step-by-step explanation:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

That is z with a pvalue of , so Z = 2.327.

Now, find the margin of error M as such

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

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​

don't know what that says sorry i can't you

step-by-step explanation:

idk what it says

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3