Question

(7.5 points) According to a recent study, the average American spent $10,345 on health care in 2016. A researcher wants to know if the amount of money residents in Louisville spend on health care is different than this national average. A survey of 300 randomly selected residents found that they spent an annual average of $9,730 on health care in 2016 with a standard deviation of $8,227. The researcher plans to run a hypothesis test at a 5% level of significance 1. (1 point) Compute the appropriate test statistic for this hypothesis test. 2. d) (1 point) Make a decision using the decision rule in (d) and the calculated test statistic in (e). Write out an appropriate conclusion. 3. e) (0.5 points) Do you expect the p-value for this test to be greater than or less than the significance level of 0.05? Explain.

Answer 1

Answer)

Null hypothesis Ho : u = 10345

Alternate hypothesis Ha : u not equal to 10345

1)

Test statistics t = (sample mean - claimed mean)/(s.d/√n)

t = (9730 - 10345)/(8227/√300) = -1.295

2)

As the population s.d is unknown here we will use t distribution

Degrees of freedom is = n-1 = 299

For 299 df and 0.05 alpha

Critical values are -1.968 and 1.968

Rejection region is

Reject Ho if test statistics is greater than 1.968 or less than -1.968

As-1.295 is not less than -1.968

We fail to reject the null hypothesis Ho

3)

as we failed to reject the null hypothesis

P-value would be greater than 0.05