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

Answers ( )

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    2021-09-11T12:59:39+00:00

    Answer:

    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 \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

    \alpha = \frac{1-0.95}{2} = 0.025

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

    So it 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*\frac{\sigma}{\sqrt{n}}

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

    M = 1.96\frac{11}{30} = 3.94

    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.

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45:7+7-4:2-5:5*4+35:2 =? ( )