Suppose you have selected a random sample of $n=14$ measurements from a normal distribution. Compare the standard normal $z$ values with the corresponding $t$ values if you were forming the following confidence intervals. $\newline$ $\begin{array}{l} \text { (a) } 98 \% \text { conf } \\ z=2.33 \\ t=2.6503 \end{array}$ $\newline$ $\text { (a) } 98 \% \text { confidence interval }$ $\newline$ (b) $99 \%$ confidence interval $\newline$ $\begin{array}{l} z=2.576 \\ t=\square \end{array}$ $\newline$ (c) $95 \%$ confidence interval $\newline$ $\begin{array}{l} z=1.96 \\ t=2.1604 \end{array}$

Full solution

Q. Suppose you have selected a random sample of

n=14

measurements from a normal distribution. Compare the standard normal

z

values with the corresponding

t

values if you were forming the following confidence intervals.

\newline

\begin{array}{l} \text { (a) } 98 \% \text { conf } \\ z=2.33 \\ t=2.6503 \end{array}

\newline

\text { (a) } 98 \% \text { confidence interval }

\newline

(b)

99 \%

confidence interval

\newline

\begin{array}{l} z=2.576 \\ t=\square \end{array}

\newline

(c)

95 \%

confidence interval

\newline

\begin{array}{l} z=1.96 \\ t=2.1604 \end{array}

Calculate t value: Calculate the t value for the $99$ % confidence interval using degrees of freedom $(n-1)$ for $n=14$ . Degrees of freedom = $14 - 1 = 13$ . Using a t-distribution table for $99$ % confidence and $13$ degrees of freedom, the t value is approximately $2.650$ .
Compare $z$ and $t$ values ( $98$ %): Compare the $z$ and $t$ values for the $98$ % confidence interval. $z$ value = $2.33$ , $t$ value = $2.6503$ .
Compare $z$ and $t$ values ( $99$ %): Compare the $z$ and $t$ values for the $99$ % confidence interval. $z$ value = $2.576$ , $t$ value = $2.650$ .
Compare $z$ and $t$ values ( $95$ %): Compare the $z$ and $t$ values for the $95$ % confidence interval. $z$ value = $1.96$ , $t$ value = $2.1604$ .