Question

6. You study a random variable that is known to be normally distributed. You obtain 21 independent measurements that yield a sample mean equal to 16.2 and a sample standard deviation equal to 3.90. Calculate the 95% confidence interval on the mean. There is a null hypothesis that the population mean is 15.0, and an altemate hypothesis that the population mean is greater than 15.0. Determine the P-value of that hypothesis test. (If it involves a Z-test, obtain the P-value using Table A4. If it involves a T-test, you will need to use Excel or some other platform to compute the P--value.)

Answer 1

a) Mean = 16.2
t critical value= 2.09
sM = √(3.9^2/21) = 0.85

μ = M ± t(sM)
μ = 16.2 ± 2.09*0.85
μ = 16.2 ± 1.775

95% CI [14.425, 17.975].

You can be 95% confident that the population mean (μ) falls between 14.425 and 17.975.

b) NULL HYPOTHESIS H0: $\mu=15$

ALTERNATIVE HYPOTHESIS Ha: $\mu>15$

alpha= 0.05

$t=16.2-15/3.9/\sqrt(21)$

$t=1.2/3.9/4.58$

$t=1.2/0.85$

$t=1.412$

degrees of freedom =n-1= 21-1=20

The P-Value is .08666.The result is not significant because p > .05.

Decision:Fail to Reject null hypothesis H0.