a) What is the p value for this test?  b) Would the null value have been rejected if this was a 1% level test? YN     c) Would the null value have been rejected if this was a 0.1% level test? YN     d) What was the value of x calculated from our sample?

MATLAB: An Introduction with Applications
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Chapter1: Starting With Matlab
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Suppose we are testing the null hypothesis H0: ? = 13 against the alternative Ha: ? > 13 from a normal population with known standard deviation ?=5. A sample of size 256 is taken. We use the usual z statistic as our test statistic. Using the sample, a z value of 2.636 is calculated. (Remember z has a standard normal distribution.)

a) What is the p value for this test? 
b) Would the null value have been rejected if this was a 1% level test?

YN    


c) Would the null value have been rejected if this was a 0.1% level test?

YN    



d) What was the value of x calculated from our sample?

Expert Solution
Step 1

a)

The hypothesis is right tail. The p-value for right tailed test is,

p-value=Pz>2.636=1-Pz<2.636

The probability of z less than 2.636 can be obtained using the excel formula “=NORM.S.DIST(2.636,TRUE)”. The probability value is 0.9958.

The required p-value is,

p-value=1-Pz<2.636=1-0.9958=0.0042

Thus, the p-value for this test is 0.0042.

Step 2

b)

Decision rule:

If p-valueα, then reject the null hypothesis. Otherwise, do not reject the null hypothesis.

Conclusion:

The p-value (0.0042) is less than the level of significance (0.01).

Based on the decision rule, reject the null hypothesis.

Yes, the value have been rejected if this was a 1% level test.

c)

Conclusion:

The p-value (0.0042) is greater than the level of significance (0.001).

Based on the decision rule, do not reject the null hypothesis.

No, the value have been not rejected if this was a 0.1% level test.

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