Solution assignment 01 of Estimation

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1. Annual income presidents
The data for this assignment are listed in the Excel file Annual income presidents.
A survey of 80 randomly selected companies asked them to report the annual income of their presidents. Assume that incomes are normally distributed with a standard deviation of 30,000 dollar.
a. Determine the 90% confidence interval estimate of the annual income of all company presidents. Interpret the statistical results.
b. What would happen to the interval width if we take the 95% confidence interval?
c. What would happen to the interval width if 250 randomly selected companies had been asked. Explain your answer.

Solution.
a. The 90% confidence interval estimate is $\displaystyle{\overline{x}\pm{z_{\alpha/2}}\frac{\sigma}{\sqrt{n}}}$ . The sample mean is $\overline{x}=$585063$ dollar (computed from the data in the data file), the population standard deviation was given (30,000 dollar), the sample size is $n=80$ and $z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645$ . This results in the interval $[579545; 590581]$ . We estimate that the mean annual income of all company presidents lies in this interval; this type of estimate is correct 90% of the time.
b. If the 95% confidence interval estimate was taken then only $z_{\alpha/2}$ changes which becomes larger: $1.96$ . The 95% confidence interval will be wider.
c. Now $n$ will be larger and thus the interval will be narrower.

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Solution assignment 01 of Estimation

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