Question

Question 26 5 pts A survey recently reported that the mean national annual expenditure for outpatient services of all persons over 64 years of age was $5,423. A random sample of 352 persons over age 64 living in Kansas had an average expense of $5,516 and a standard deviation of $979. We want to test if the average outpatient expenditure (mu) is higher in Kansas (compared to national average). What is the p-value for the test? 0.099 0.037 0.963 0.010 0.074

Answer 1

this is the two tailed test .  

The null and alternative hypothesis is ,

H0 :  $\mu$ = 5423

Ha : $\mu$    $\neq$ 5423

Test statistic = z

= ($\bar x$ - $\mu$ ) / $\sigma$ / $\sqrt$ n

= (5516-5423) / 979 / $\sqrt$ 352

= 1.78

P(z > 1.78 ) = 1 - P(z < 1.78 ) = 1- 0.9625=0.0375

P-value = 2*0.0375=0.074