Question

The mean and standard deviation of a population are 11,795 and 14054 respectively. If n=50 ,find 95% confidence
interval for the mean.

Answer 1

Answer:

Step-by-step explanation:

Here mean of population , mu = 11795

S.D of population , sigma = 1405

x bar = 11795

n = sample size = 50 , maximum error = Z alpha by 2 x sigma by root n

Z alpha by 2 for 95%  confidence = 1.96

Max. error, E = Z alpha by 2 x sigma/ root n = 1.96 x 14054/ root of(50) = 3899

Therefore,  Confidence interval = ( x bar - z alpha by 2 . sigma / root n , x bar + Z alpha by 2 . sigma / root n

=(11795 - 3899, 11795 + 3899)

=(7896, 15694)

Answer 2

Given, the mean of the  population $\mu = 11795$

the standard deviation $\sigma = 14054$

Sample size $n = 50$

Maximum error $E = Z_{\alpha/2}\frac{\sigma}{\sqrt{n} }$

Now,Critical value $Z_{\alpha/2}$ for 95% confidence = 1.96

$\therefore E = 1.96\frac{14054}{\sqrt{50} }= 3893.075$

Confidence Interval = $(\bar{x}-z \frac{\sigma}{\sqrt{n}}) to (\bar{x}+z \frac{\sigma}{\sqrt{n}})$

                                 =$(11795-3893.075) to (11795+3893.075)$

                                 = $(7901.92,15688.08)$

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