Question

An important property of plastic clays is the percent of shrinkage on drying. A certain type of plastic clay has a shrinkage that has a standard deviation of 1.2 percent. A random sample of 38 results in an average shrinkage of 18.4 percent. Test to determine whether the true average shrinkage is greater than 18 percent. Use alpha = .05

Answer 1

Claim : Test to determine whether the true average shrinkage is greater than 18 percent.

1 ) Hypothesis :

Null hypothesis : the true average shrinkage is less than or equal to 18 percent.

Alternative hypothesis : the true average shrinkage is greater than 18 percent.

2 ) Rejection region :

Here alternative hypothesis contain " > " sign since test is right tailed test .

So here to find critical calue use area as , 1-0.05 = 0.95 .

with excel command , =NORMSINV(0.95) we get p- value as 1.645 .

The rejection region for this right-tailed test is R = { z test statitics value > 1.645}

3) Test Statistics :

4) Decision about the null hypothesis:

So here Z test statitics value =2.055 > 1.645 ;it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the true average shrinkage is greater than 18 percent at the 0.05 significance level.