Question

It is believed that the average amount of money spent per U.S. household per week on food is about $98, with standard deviation $10. A random sample of 100 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. What is the test statistic and p-value for the problem

Answer 1

Answer:

Test statistic = 2

P-value = 0.0227                                

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $98

Sample mean, $\bar{x}$ = $100

Sample size, n = 100

Population standard deviation, σ = $10

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 98\text{ dollars}\\H_A: \mu > 98\text{ dollars}$

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{100 - 98}{\frac{10}{\sqrt{100}} } = 2$

Now, we can calculate the p-value from the normal table

P-value = 0.0227