Question

58. Critical values What critical value t* from Table$B$ should be used for a confidence interval for the population mean in each of the following situations?
(a) A $90\%$confidence interval based on n  $n=12$ observations.
(b) A $95\%$confidence interval from an SRS of$30$observations.

Short answer

(a) A critical value for $90\%$confidence interval based on $n=12$observations is $t^{*}=1.796$

(b) A critical value for $95\%$confidence interval from an SRS of $30$observations is $t^{*}=2.045$.

Step-by-step solution

Part (a) Step 1: Given information 

When the confidence level is $90\%$and the population mean is $n=12$, determine the critical value $t*$.

Part (a) Step 2: Explanation 

Using$df=n-1$, to determine the degree of freedom .

Where, the population mean $n$is$12$.
$$\begin{gathered} df=n-1 \\ =12-1 \\ =11 \end{gathered}$$
Convert the confidence level$90\%$ into decimal:
$\frac{90}{100}=0.90$
Determine the column:
$$\begin{gathered} \frac{1-c}{2}=\frac{1-0.90}{2} \\ =0.05 \end{gathered}$$
From table $B$, determine the critical value $t^{*}$, for row $11$and the column$0.05$:
$t^{*}=1.796$
Therefore, the critical value is $t^{*}=1.796$.

Part (b) Step 1: Given information 

When the confidence level is $95\%$and the population mean is $30$, determine the critical value $t^{*}$.

Part (b) Step 2: Explanation 

Using $df=n-1$, to determine the degree of freedom, where the population mean $n=30.$
$df=n-1=30-1=29$
From the table $B$, the degree of freedom$df$ indicates the row .
Convert the confidence level$95\%$into decimal:
$\frac{95}{100}=0.95$
Determine the column:
$$\begin{gathered} \frac{1-c}{2}=\frac{1-0.95}{2} \\ =0.025 \end{gathered}$$

From the table $B,$the critical value $t^{*}$, for row$9$and the column $0.025$:

$t^{*}=2.045$.

Therefore, the critical value is $2.045$.