Recess and Wasted Food. Refer to Exercise 10.50 and find a 98% confidence interval for the difference between the mean amount of food wasted for lunches before recess and that for lunches after recess.

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:

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

Then consider the lunch after recess,

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

The null and alternate hypotheses should be stated.

Null hypotheses: $H_0:{\mu }_1\le {\mu }_2$

Alternate hypotheses: $H_a:{\mu }_1>{\mu }_2$

Hypotheses have a right-tailed distribution.

Use Table IV to find $t_{\alpha /2}$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 $=\left({\overline{x}}_1-{\overline{x}}_2\right)\pm t_{\alpha /2}·s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}$

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

Confidence interval $=66.5\pm 12.0$

Confidence interval $=54.5to78.5$