Question

1. According to AARP, in 2008, 49% of all annual expenditure on restaurant food was by Americans age 50+. In fact AARP claims the average annual expenditure for Americans age 50+ on restaurant food in 2008 was $1960. Suppose a 2015 study randomly sampled 42 Americans age 50+ and found an average annual expenditure on restaurant food of $2145 with a standard deviation of $600. Is there reason to believe that the average annual expenditure for Americans age 50+ on restaurant food has increased since 2008 at α=.025?

   For the hypothesis stated above, what is the decision?

   	a.	Reject H0 because the test statistic is to the right of the positive critical value
   	b.	Fail to reject H0 because the test statistic is to the right of the positive critical value
   	c.	Fail to reject H0 because P-value > α
   	d.	None of the answers is correct
   	e.	Reject H0 because P-value > α

Answer 1

Step 1:

Ho:

Ha:

Null hypothesis states that the average annual expenditure for americans age 50+ on resturant food is 1960.

Step 2:

n = 42

sample mean = 2145

sample sd = 600

Assuming that the data is normally distributed, also as the population sd is not given, we will calculate t stat

p value using excel formula

P value = TDIST (t statistics, df, 1) = TDIST(1.998,41,1)= 0.0262

P value = 0.0262

Step 3:

The t-critical value for a right-tailed test, for a significance level of α=0.025

tc=2.02

As the t stat does not fall in the rejection area we fail to reject the Null hypothesis.

Also as the p value 0.0262 is greater than level of significane (0.025) we fail to reject the Null hypothesis.

Option C : Fail to reject H0 because P-value > α