Tazmin Horton

2020-12-15

You are given the sample mean and the population standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals. From a random sample of 58 ​dates, the mean record high daily temperature in a certain city has a mean of ${83.43}^{\circ }F$. Assume the population standard deviation is ${14.02}^{\circ }F$. $90\mathrm{%}=$ $95\mathrm{%}=$ Which interval is wider?

Obiajulu

Step 1 Given Data: The sample size is: $n=58$ The mean temperature of the city is: $\mu ={83.43}^{\circ }F$ The standard deviation of the temperature is: $\sigma ={14.02}^{\circ }F$ 95% Population: The expression calculate the 95% of the Population is, $1-2\alpha =0.95$
$\alpha =\frac{1-0.95}{2}$
$=0.025$ Corresponds to probability 0.025, the z value from the right tail end of the z-table 1.96. Step 2 The expression to calculate the interval of the population is, $CI=\stackrel{―}{X}±M.E$
$=\stackrel{―}{X}±Z×\frac{\sigma }{\sqrt{n}}$ Substitute values in the above expression. $CI=83.43±1.96×\frac{14.02}{\sqrt{58}}$
$=83.43±3.60$
$=\left(83.43-3.60,83.43+3.60\right)$
$\left(79.83,87.03\right)$ Thus, the interval width of the 95% level is 3.6. and the interval is (79.83, 87.03). Step 3 90% Population: The expression calculate the 95% of the Population is, $1-2\alpha =0.90$
$\alpha =\frac{1-0.90}{2}$
$=0.050$ Corresponds to probability 0.050, the z value from the right tail end of the z-table is 1.64. The expression to calculate the interval of the Population is, $CI=\stackrel{―}{X}±M.E$
$=\stackrel{―}{X}±Z×\frac{\sigma }{\sqrt{n}}$ Substitute values in the above expression. $CI=83.43±1.96×\frac{14.02}{\sqrt{58}}$
$=83.43±3.02$
$=\left(83.43-3.60,83.43+3.02\right)$
$=\left(80.41,86.45\right)$ Thus, the interval width of 90% population level is 3.02, and the interval is (80.41, 59.45). Thus, 95% population has higher interval width.

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