P3 – where applicable, readings from tables must be used.
The quality-control manager at a light bulb factory needs to determine whether the mean life of a large shipment of light bulbs is equal to 375 hours. The population standard deviation is 100 hours. A random sample of 64 light bulbs indicates a sample mean life of 350 hours.
a)At the 0.05 level of significance, is there evidence that the mean life is different from 375 hours?
b)Compute the p-value and interpret its meaning.
c)Construct a 95% confidence interval estimate of the pupulation mean life of the light bulbs.
d)Compare the reuslts of (a) and (c). What conclusions do you reach?
2.If, in a sample of n = 16 selected from a normal population, X ̅ = 56 and S = 12, what is the value of tSTAT if you are testing the null hypothesis H0: = 50?
3.In problem 2 above, how many degrees of freedom are there in the t- test?
4.In problems 2 and 3 , what are the critical values of t if the level of significance, α is 0.05 and the alternative hypothesis, H1, is ≠ 50?
5.In problem 2, 3 and 4, what is your statistical decision if the alternative hypothesis, H1 is ≠ 50?
6.In a training process, the average time taken is 6.4 hours. Eight employees were trained using a new method and they had an average training time of 6.2 hours and a standard deviation of 1.1 hours. Use α = 0.01 to determine if the new process reduced the training time.
In testing for differences between the means of two related populations, the null hypothesis is
H0 : D = 2.
H0 : D = 0.
H0 : D < 0.
H0 : D > 0.
A powerful women’s group has claimed that men and women differ in attitudes about sexual discrimination. A group of 50 men (group 1) and 40 women (group 2) were asked if they thought sexual discrimination is a problem in the United States. Of those sampled, 11 of the men and 19 of the women did believe that sexual discrimination is a problem. Find the value of the test statistic.
Z = -2.55
Z = -0.85
Z = -1.05
Z = -1.20
When testing H0 : π 1 – π 2 0 versus H1 : π 1 – π 2 > 0, the observed value of the Z-score was found to be -2.13. The p-value for this test would be
Given the following information, calculate the degrees of freedom that should be used in the pooled-variance t test.
s12 = 4 s22 = 6
n1 = 16n2 = 25
df = 41
df = 39
df = 16
df = 25