Chapter 9, Problem T9.13SPT

To determine

To Explain: that there is convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158.

Answer to Problem T9.13SPT

There is enough proof to help the claim that the mean weekly spending on food in this city differs from the national figure of $158.

Explanation of Solution

Given:

  $\begin{array}{c}n=50\\ \overline{x}=165.00\\ s=20.00\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

The hypotheses are

  $\begin{array}{c}{H}_0:\mu =\$158\\ H_1:\mu \ne \$158\end{array}$

The value of the test statistic

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{165.00-158}{20.00/\sqrt{50}}\\ =2.475\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Table B containing the t-value in the row $n-1=50-1=49>40:$

  $0.01=2\times 0.005<P<2\times 0.01=0.02$

If the P-value is lesser than the significance level, then the null hypothesis is rejected.

  $P<0.05=5\%\Rightarrow \text{Reject }{H}_0$

There is enough proof to help the claim that the mean weekly spending on food in this city differs from the national figure of $158.