Question

A random sample of 16 values from a normal population showed a mean of 41.5 inches

and sum of square deviation from the mean equal to 135 inches. Obtain 95% and 99% confidence

interval for the population mean?​

Answer 1

Step-by-step explanation:

we conclude that the data are consistent with the assumption of mean I.Q of 100 in the population. Problem:2 A random sample of 16 values from a normal population showed a mean of 41.5 inches and the sum of squares of deviations from this mean is equal to 135 Square inches.

Answer 2

The assumption of a mean of 43.5 is NOT reasonable at 95% confidence interval as it does not fall within the interval range but it is reasonable at 99% confidence interval as it fall within the interval range.

Given that,

The mean value was 41.5 inches, and the sum of square deviations from the mean was 135 inches, according to a randomly selected sample of 16 values from a normal population.

We have to obtain the population mean's 95% and 99% confidence intervals.

We know that,

$\bar{x}$  =41.5

s=$\sqrt{\frac{135}{16-1} }$ =3

n = 16

Confidence interval:

CI = $\bar{x}$  ± Margin of error (E)

Standard error sx=$\frac{s}{\sqrt{n} }$

df= n-1=15

Critical value for 95% confidence interval tc​ =2.49

CI = 41.5 ± 2.490 × 0.75

CI = 41.5 ± 1.8674

CI = (39.632,43.367)

Critical value for 99% confidence interval tc =3.286

CI = 41.5 ± 3.286 × 0.75

CI = 41.5 ± 2.4645

CI = (39.035,43.964)

Therefore, The assumption of a mean of 43.5 is NOT reasonable at 95% confidence interval as it does not fall within the interval range but it is reasonable at 99% confidence interval as it fall within the interval range.

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