91Ó°ÊÓ

Lunch before recess and lunch after recess are two populations from which sample data is available.

The significance level is 1%.

Let's consider the lunch before recess:

$$\begin{gathered} {\stackrel{¯}{x}}_1=223.1, \\ {s}_1=122.9, \\ \text{and} \\ {n}_1=889 \end{gathered}$$

Then consider the lunch after recess,

The main goal is to determine a $95\%$confidence interval for the difference between two population mean ${\mathrm{μ}}_1$and ${\mathrm{μ}}_2$

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0:{\mathrm{μ}}_1≤{\mathrm{μ}}_2$

Alternate hypotheses: $H_a:{\mathrm{μ}}_1>{\mathrm{μ}}_2$

Hypotheses have a right-tailed distribution.

Use Table IV to find $t_{\mathrm{Î\pm }/2}$with $\mathrm{df}=n_1+n_2-2$for a confidence level of $1-a$.

alpha=0.02, for 98% confidence level,

localid="1651406295916" $$\begin{gathered} \text{df}=n_1+n_2-2 \\ =(889+1119-2) \\ =2006. \end{gathered}$$

When $\mathrm{df}=2006$, using table IV for critical values,

Critical value,

Determine the confidence interval's endpoints.

Pooled standard deviation,

$S_p=\sqrt{\frac{(889-1)(122.9{)}^2+(1119-1\left)\right(108.1{)}^2}{889+1119-2}}$

$s_p=\sqrt{\frac{888(122.9{)}^2+1118(108.1{)}^2}{2006}}$

$s_p=114.887$

Confidence interval $=\left({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2\right)Â\pm t_{\mathrm{Î\pm }/2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

Confidence interval $=(223.1-156.6)Â\pm 2.328×114.887\sqrt{\frac{1}{889}+\frac{1}{1119}}$

Confidence interval $=66.5Â\pm 12.0$

Confidence interval $=54.5to78.5$