Question

A random sample of size 16 has 53 as mean. The sum of the squares of the
deviations taken from the mean is 150. Can this sample be regarded as taken
from the population having 56 as mean? Obtain 95% and 99% confidence limits
of the mean population. (For v =15, to.o. -2.95, to.os-2.131).
​

Answer 1

A random sample of size 16 has 53 as mean. The sum of the squares of the

deviations taken from the mean is 150. Can this sample be regarded as taken

from the population having 56 as mean? Obtain 95% and 99% confidence limits

Answer 2

The population having 56 as mean, as respectively 51.316< x̄ < 54.684 and 50.67< x̄ < 55.33, Obtain 95% and 99% confidence limits.

Given,

N= 16,

x̄= 53,

μ= 56

To Find,

We have to find, Can this sample be regarded as taken from the population having 56 as mean? Obtain 95% and 99% confidence limits

Solution,

S= √ Σ(X-x̄)/N-1

or, S= √150/15 = √10= 3.162

t computed is, t= (x̄- μ)/S √N

or, t= |53-56|√16/ 3.162

or, t= 3.795

t critical is; df= 16−1 = 15

α=0.05

t (critical) =2.131

So, Since t_{computed}>t_{critical},the result of the experiment does not support the hypothesis that the sample is taken from the universe having a mean 56.

95% confidence limit ;

x̄±S/√N t0. 05 = 53±3.162/√16 *2.131

= 53±1.6846

51.316< x̄ < 54.684

99% confidence limit;

x̄±S/√N t0. 01= 53±3.162/√16* 2.947

= 53± 2.330

50.67< x̄ < 55.33

Hence, The population having 56 as mean, as respectively 51.316< x̄ < 54.684 and 50.67< x̄ < 55.33, Obtain 95% and 99% confidence limits.

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