Q.13 - The Practice of Statistics for AP

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 Minitab output below shows the results of requesting a confidence interval for the population mean M. An examination of the data reveals no outliers.

(a) Explain why the Normal condition is met in this case. 

(b) Can you conclude that the mean weekly spending on food in this city differs from the national figure of $\backslash )158$? Give appropriate evidence to support your answer. 

Step-by-Step Solution

Answer

a).  The Normal requirement has been satisfied.

b). There is sufficient evidence to support the claim that the mean weekly spending on food in this city differs from the national figure of $\$158$.

Part (a) Step 1: Given Information

Part (a) Step 2: Explanation

The Normal requirement requires no outliers and that the distribution is approximately normally distributed.

Given is that the data set contains no outliers.

The sample size of $50$ is $30$ or more and thus we can assume that the distribution is approximately normal.

Thus the Normal requirement has been satisfied.

Part (b) Step 1: Given Information

$n=50$

$\overline{x}=165.00$

$s=20.00$

Part (b) Step 2: Explanation

Determine the hypotheses:

$H_0:\mu =\$158$

$H_0:\mu \ne \$158$

Determine the value of the test statistic:

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

$=2.475$

The $p$-value is the likelihood of getting the test statistic's value or a number that is more severe. The $p$-value is the number (or interval) in Table B's column header for the $t$-value in row  $n-1=50-1$ 

        $=49>40$:

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

        $=0.02$

The null hypothesis is rejected if the $P$-value is less than the significance level.

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