Question

The following sample of 16 measurements was selected from a population that is approximately normally distributed:

1. Construct an 80% confidence interval for the population mean.
2. Construct a 95% confidence interval for the population mean and compare the width of this interval with that of part a.
3. Carefully interpret each of the confidence intervals and explain why the 80% confidence interval is narrower.

Short answer

A confidence interval is described as the set of numbers observed in our collection for which we anticipate discovering the figure that best represents the whole population.

Step-by-step solution

(a) The data is given below

The calculation is given below:

$$\begin{gathered} \text{n = 16} \\ \overline{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$

$$\begin{gathered} \text{df = 16}-\text{1} \\ \text{= 15} \end{gathered}$$

$\alpha =0.20$

The critical worth of localid="1652095550050" $t=\pm \text{1.341}$

The confidence interval is 80%:

localid="1652095596149" $\overline{X}\frac{t\times s}{\sqrt{n}}=97.9375\pm \frac{1.341\times 12.6463}{\sqrt{16}}=97.9375\pm 4.238=\left(93.699,102.176\right)$

$\text{The}80\%\text{confidential level}\left(93.699,102.176\right)$

(b) The data is given below

The calculation is given below:

$$\begin{gathered} \text{n = 16} \\ \overline{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$$$\begin{gathered} \text{n = 16} \\ \overline{X}=\text{97.9375} \\ \text{s = 12.6463} \end{gathered}$$

$$\begin{gathered} \text{df = 16}-\text{1} \\ \text{= 15} \end{gathered}$$

$\alpha =0.20$

The critical worth of $t=\pm \text{1.341}$

The confidence interval is 80%:

$\overline{X}\frac{t\times s}{\sqrt{n}}=97.9375\pm \frac{2.131\times 12.6463}{\sqrt{16}}=97.9375\pm 6.739=\left(91.199,104.676\right)$

$\text{The}95\%\text{confidential level}\left(91.199,104.676\right)$

The confidence interval 80% is (93,699,102.176) is narrower than confidence interval 95% is (91.199, 104.676)

(c) The data is given below

As the level of confidence goes from 95% to 80%, the width of the confidence interval narrows. Since lowering the confidence threshold from 95% to 80% reduces the error margin, resulting in a smaller confidence interval. As the level of confidence lowers, so does our estimation of the genuine unknown population parameter to be around a specific interval.