Question

In each of the $8.123-8.128$Exercises \(8.123-8.128\), we provide a sample mean, sample size, sample standard deviation, and confidence level. In each exercise,

a. use the one-mean t-interval procedure to find a confidence interval for the mean of the population from which the sample was drawn.

b. obtain the margin of error by taking half the length of the confidence interval.

c. obtain the margin of error by using the formula $t_{u/2}·3/\sqrt{n}$.

$\overline{x}=30,n=25,s=4$, confidence level $=90\%$

Short answer

Part (a) The $90\%$confidence interval for $\mu$is $(28.6312,31.3688)$

Part (b) The margin of error by using the half-length of the confidence interval is $1.3688$

Part (c) The margin of error by using the formula is $1.3688$

Step-by-step solution

Part (a) Step 1: Given information

$\overline{x}=30,n=25,s=4$, confidence level $=90\%$

Part (a) Step 2: Concept

The formula used: The confidence interval$\overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)$and$\text{Margin of error}\left(E\right)=t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)$

Part (a) Step 3: Calculation

Calculate the $90\%$confidence interval for $\overline{x}=30,n=25,s=4$and the requisite degrees of freedom are$1.711$from "Table IV Values of $t_{\alpha }$". Thus, the $90\%$confidence interval is,

$$\begin{gathered} \overline{x}\pm t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right)=30\pm 1.711\left(\frac{4}{\sqrt{25}}\right) \\ =30\pm 1.3688 \\ =(30-1.3688,30+1.3688) \\ =(28.6312,31.3688) \end{gathered}$$

Therefore, the $90\%$confidence interval $\mu$is $(28.6312,31.3688)$

Part (b) Step 1: Calculation

Using the half length of the confidence interval, calculate the margin of error.

$$\begin{gathered} \text{Margin of error}=\frac{\text{Upper limit}-\text{Lower limit}}{2} \\ =\frac{31.3688-28.6312}{2} \\ =\frac{2.7376}{2} \\ =1.3688 \end{gathered}$$

Thus, the margin of error by using the half-length of the confidence interval is $1.3688$

Part (c) Step 1: Calculation

Using the formula, calculate the margin of error.

$$\begin{gathered} \text{Margin of error}\left(E\right)=t_{\frac{a}{2}}\left(\frac{s}{\sqrt{n}}\right) \\ =1.711\left(\frac{4}{\sqrt{25}}\right) \\ =1.3688 \end{gathered}$$

Thus, the margin of error by using the formula is $1.3688$