Question

assume the mean forced expiratory volume (FEV1) is 1.08 liters. A random sample of 55, 3 to 6-year-olds who live in a community with high level of ozone pollution are found to have a sample mean FEV1 of 0.99 L with standard deviation of 0.28 L. Use the 5% significance level to test the claim that young children from this high population community have lower than average. what is the p-value?

Answer 1

Solution :

Given that

= 1.08

= 0.99

= 0.28

n = 55

This is the left tailed test .

The null and alternative hypothesis is ,

H0 :   = 1.08

Ha : < 1.08

Test statistic = z

= ( - ) / / n

= ( 0.99 - 1.08 ) / 0.28 / 55

= -2.38

The test statistic = -2.38

P ( Z < -2.38 ) = 0.0087

The P-value = 0.0087