Question

A government report says that the average amount of money spent per U.S. household per week on food is about $\backslash (158$. A random sample of$50$ households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of \)165 and a standard deviation of $\backslash (20$. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $\backslash )158$?

Short answer

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

Step-by-step solution

Step 1:Given information

$n=50$

$\overline{x}=165.00$

$s=20.00$

Step 2:Explaination

The hypotheses are

$H_0:\mu =\$158$

$H_1:\mu \ne \$158$

The value of the test statistic

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

$=\frac{165.00-158}{20.00/\sqrt{50}}$

$=2.475$

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$\$158$