## 1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample: 43 52 18 20 25 45 43 21 42 32 24 32 19 25 26

Question

1) Find the 95 % confidence interval for the mean if σ = 11 , using this sample:
43 52 18 20 25 45 43 21 42 32 24 32
19 25 26 44 42 41 53 22 25 23 21 27
33 36 47 19 20 41

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1 month 2021-09-11T12:58:26+00:00 1 Answer 0

The 95 % confidence interval for the mean is between 28.09 and 35.97.

Step-by-step explanation:

Mean of the sample:

30 values.

So we sum all the values and divide by 30.

The sum of all the values(43 + 52 + 18 + … + 20 + 41) is 961.

961/30 = 32.03

Confidence interval:

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 .

So it is z with a pvalue of , so 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 32.03 – 3.94 = 28.09

The upper end of the interval is the sample mean added to M. So it is 32.03 + 3.94 = 35.97

The 95 % confidence interval for the mean is between 28.09 and 35.97.