Question

construct the confidence interval for the population mean $mu$.
$c = 0.98,\bar{x}=4.6,sigma = 0.5$, and $n = 43$
a 98% confidence interval for $mu$ is ( ). (round to two decimal places as needed.)

Explanation:

Step1: Find the z - value

The confidence level $c = 0.98$. The significance level $\alpha=1 - c=1 - 0.98 = 0.02$. Then $\frac{\alpha}{2}=0.01$. Looking up in the standard - normal distribution table, the $z$ - value $z_{\alpha/2}=z_{0.01}\approx 2.33$.

Step2: Calculate the margin of error $E$

The formula for the margin of error when the population standard deviation $\sigma$ is known is $E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Substitute $z_{\alpha/2}=2.33$, $\sigma = 0.5$, and $n = 43$ into the formula. First, calculate $\sqrt{n}=\sqrt{43}\approx6.56$. Then $\frac{\sigma}{\sqrt{n}}=\frac{0.5}{6.56}\approx0.076$. And $E = 2.33\times0.076\approx0.18$.

Step3: Construct the confidence interval

The confidence interval for the population mean $\mu$ is given by $\bar{x}-E<\mu<\bar{x} + E$. Substitute $\bar{x}=4.6$ and $E = 0.18$ into the formula. We get $4.6-0.18<\mu<4.6 + 0.18$, which is $4.42<\mu<4.78$.

Answer:

$(4.42,4.78)$