Question

Hi, can you help me answer this question please, thank you

Answer 1

From the given data;

`\begin{gathered} \text{ sample size, n=43} \\ \text{ sample mean, }\bar{x}=64 \\ \text{ population standard deviation,}\sigma=14.1 \end{gathered}`
`\begin{gathered} \text{significance level,}\alpha=1-confidence\text{ interval} \\ \alpha=1-0.9 \\ \alpha=0.1 \end{gathered}`
`The\text{ critical value}=z_{(\alpha)/(2)}=z_{(0.1)/(2)}=z_(0.05)=1.645\text{ ( from the }z-table)`
`\begin{gathered} \text{Therefore,} \\ critical\text{ value}=\pm z_{(\alpha)/(2)}=\pm1.645 \end{gathered}`
`\text{Margin of error, E=}z_{(\alpha)/(2)}*\frac{\sigma}{\sqrt[]{n}}`
`\begin{gathered} E=1.645*\frac{14.1}{\sqrt[]{43}} \\ E=1.645*2.1502 \\ E=3.5371 \end{gathered}`

Limits of 90% confidence interval are given by:

`\begin{gathered} \text{lower limit}=\bar{x}-E \\ \text{lower l}imit=64-3.5371 \\ \text{lower limit=60.463} \end{gathered}`
`\begin{gathered} upper\text{ limit=}\bar{\text{x}}+E \\ \text{upper limit=64+3.5371} \\ \text{upper limit=67.5371} \end{gathered}`

Hence, the 90% confidence interval for the true population mean textbook weight is:

`60.463<\mu<67.5371`