Question

Andrew needs to weigh less than 199.0 lbs to meet the body mass index requirement for his employer's health insurance discount, so he has been monitoring his weight on his scale at home. According to the manufacturer's website, Andrew's scale has measurement error, and the errors are normally distributed with a standard deviation of 1.2 lbs. To account for this error, he selected a random sample from his last month of measurements and recorded his summary statistics in the following table. test of ?=199.0 vs ?<199.0 the assumed standard deviation=1.2

Sample size n=10

Sample mean 198.5

Standard error .37947

Complete the analysis by calculating the p- value and making the decision. Give the p- value precise to at least four decimal places.

p=

Answer 1

Solution :

= 199

= 198.5

= 1.2

n =10

This is the left tailed test .

The null and alternative hypothesis is ,

H₀ :   = 199

H_a : < 199

Test statistic = z

= ( - ) / / n

= (198.5-199) / 1.2 / 10

= -1.318

P(z < -1.318) = 0.1101

P-value = 0.1101