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_{\frac{\alpha}{2}}=z_{\frac{0.1}{2}}=z_{0.05}=1.645\text{ ( from the }z-table)$$\begin{gathered} \text{Therefore,} \\ critical\text{ value}=\pm z_{\frac{\alpha}{2}}=\pm1.645 \end{gathered}$$\text{Margin of error, E=}z_{\frac{\alpha}{2}}\times\frac{\sigma}{\sqrt[]{n}}$$\begin{gathered} E=1.645\times\frac{14.1}{\sqrt[]{43}} \\ E=1.645\times2.1502 \\ 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:

$60.463<\mu<67.5371$