 # A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is \sigma=15 Compute the 95\% confidence interval for the population mean. UkusakazaL 2021-08-03 Answered
A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is $\sigma =15$
a) Compute the $95\mathrm{%}$ confidence interval for the population mean. Round your answers to one decimal place.
b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a $95\mathrm{%}$ confidence interval for the population mean. Round your answers to two decimal places.
c) What is the effect of a larger sample size on the interval estimate?
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a) The sample size is 60, sample mean is 80, and population standard deviation is 15.
Computation of critical value:
$\left(1-\alpha \right)=0.95$
$\alpha =0.05$
$\frac{\alpha }{2}=0.025$
$1-\frac{\alpha }{2}=0.975$
The critical value of z-distribution can be obtained using the excel formula $“=NORM.S.INV\left(0.975\right)”$. The critical value is 1.96.
The margin of error is,
$E={z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$
$=1.96\left(\frac{15}{\sqrt{60}}\right)$
$=3.80$
The margin of error is 3.80.
The $95\mathrm{%}$ confidence interval for the population mean is,
$C.I.=\stackrel{―}{x}±E$
$=80±3.80$

Thus, the $95\mathrm{%}$ confidence interval for the population mean is .
b) The margin of error is,
$E={z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)$
$=1.96\left(\frac{15}{\sqrt{120}}\right)$
$=2.68$
The margin of error is 2.68.
The $95\mathrm{%}$ confidence interval for the population mean is,
$C.I.=\stackrel{―}{x}±E$
$=80±2.68$

Thus, the $95\mathrm{%}$ confidence interval for the population mean is .
c) The margin of error for sample size 60 is 3.80, and margin of error for sample size 120 is 2.68. It is clear that increasing the sample size causes the decreases the margin of error value
Thus, the larger sample provides a smaller margin of error.