Chapter 9, Problem 13PT

a.

To determine

To explain:The normal condition is met in this case.

Answer to Problem 13PT

The sample size is more than 30 and given that there are no outliers. So the Normal conditionis met.

Explanation of Solution

Given information:

The average amount of money spent per U.S. household per week on food = $\text{\$}158$ .

Number of random sample of households in a small city = 50

  

The sample size is more than 30 and given that there are no outliers. So the Normal conditionis met.

(b)

To determine

The mean weekly spending of food in the city differs from the national figure of $\text{\$}158$ .

Answer to Problem 13PT

The null hypothesis is rejected, because the interval does not contain mean. So it is concluded that the mean weekly spending on food in the city differs from the national figure of $\text{\$}158$ .

Explanation of Solution

To test the claim that the mean weekly spending on food in the city differs from the

national figure of $\text{\$}158$

That is $H_0:\mu =158$ versus $H_a:\mu \ne 158$

Plan:

Using one sample $t$ interval for $\mu$ if the conditions are satisfied.

1) Random: Select a random sample of 50 households and their weekly spending is recorded.

2) Normal: Given that there are no outliers. Also the sample size $n=50$ , is large (at least 30 ), so we are safe to use $t$ procedures.

3) Independent: To check the independence, first check the $10\text{\%}$ condition

It is known that there will be at least 500 households. So the $10\text{\%}$ condition is satisfied.

Given the $95\text{\%}$ confidence interval of $\mu$ is given $\overline{x}\pm t^{\ast }\frac{s_x}{\sqrt{n}}=(159.32,170.68)$

The null hypothesis is rejected, because the interval does not contain mean. So it is concluded that the mean weekly spending on food in the city differs from the national figure of $\text{\$}158$ .