Question

A random sample of forty-two 400-meter swims has a mean time of 3.18 minutes and a standard deviation of 0.07 minutes. Construct a 90% confidence interval for the population mean time.3.16 < μ < 3.203.14 < μ < 3.223.12 < μ < 3.243.18 < μ < 3.19

Answer 1

`\begin{gathered} \text{Given} \\ \overline{x}=3.18 \\ s=0.07 \\ n=42 \end{gathered}`

Recall that the formula for confidence interval is determined by

`\begin{gathered} CI=\bar{x}\pm z\frac{s}{\sqrt[]{n}} \\ \text{where} \\ \bar{x}\text{ is the sample mean} \\ s\text{ is the sample standard deviation} \\ n\text{ is the number of data} \\ z\text{ is the z-score of the confidence interval} \end{gathered}`

A 90% confidence interval has a z-score of 1.645. Substitute the following values and we get

`\begin{gathered} CI=\bar{x}\pm z\frac{s}{\sqrt[]{n}} \\ CI=3.18\pm(1.645)\frac{0.07}{\sqrt[]{42}} \end{gathered}`

Solve for the upper and lower limit

`\begin{gathered} \text{Upper Limit} \\ 3.18+(1.645)\cdot(0.07)/(√(42))\approx3.20 \\ \\ \text{Lower Limit} \\ 3.18-(1.645)\cdot\frac{0.07}{\sqrt[]{42}}\approx3.16 \end{gathered}`

Therefore, the 90% confidence interval for the population mean time is

`3.16<\mu<3.2`