Question

(b) A random sample of size 16 values from a normal
population showed a mean of 53 and a sum of squares
of deviation from the mean equals to 150. Can this
sample be regarded as taken from the population
having 56 as mean ? Obtain 95% and 99% confidence
limits of the mean of the population.
(7)​

Answer 1

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Answer 2

The population having 56 as mean, as respectively $51.316 < \bar{X} < 54.684$ and $50.67 < \bar{X} < 55.33$ , Obtain 95% and 99% confidence limits.

Step-by-step explanation:

Given:

A random sample of size 16 values from a normal population.

A mean of  53.

A sum of squares of deviation from the mean equals to 150.

To Find:

This sample  can be regarded as taken from the population having 56 as mean.  Obtain 95% and 99% confidence limits of the mean of the population.

Solution:

A random sample of size values from a normal population,N=16

Mean,$\bar{x}=53.$

A sum of squares of deviation from the mean  $\mu$ = 150.

$S=\sqrt{\sum\frac{(X-\bar{x})}{N-1} }$

    $=\sqrt{\frac{150}{16-1} } =\sqrt{\frac{150}{15} }$

    $=\sqrt{10} = 3.162$

t computed,

$\ t=\frac{(\bar{X}-\mu)}{S }\times \sqrt{N}$

  $=\frac{|53-56|}{16 }\times\sqrt{16}$

$t= 3.795$

t critical,  

$df= 16-1 = 15$

$\alpha =0.05$

t critical =2.131

Therefore,

Because $t_{computed} > t_{critical}$, 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

$\bar{X}\±\frac{S}{\sqrt{N} } t_{0.05}=53}\±\frac{3.162}{\sqrt{16} } t_{2.131}$

                     $=53\±1.6846$

$51.316 < \bar{X} < 54.684$

99% confidence limit

$\bar{X}\±\frac{S}{\sqrt{N} } t_{0.01}=\ 53}\±\frac{3.162}{\sqrt{16} } t_{2.947}$

                     $=53\±2.330$    

$50.67 < \bar{X} < 55.33$              

Thus, the population having 56 as mean, as respectively $51.316 < \bar{X} < 54.684$ and $50.67 < \bar{X} < 55.33$ , Obtain 95% and 99% confidence limits.

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