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1. Imagine that the population mean for the number of books read by 3rd graders is 47
with a population standard deviation of 8. The third grade class at Southview Elementary
initiated a reading incentive program and based on a sample of 150 children, obtained a
mean of 55 books. The research question is whether the reading incentive program was
successful in increasing the number of books read by 3rd graders. What test would be
used? Calculate the test statistic.
The z- test for a population mean is used.
Here it is given that, m = 47, σ = 8, x-bar = 55, n = 150
The test statistic used is
z = (x-bar - m)/( σ /sqrt(n))
= (55 - 47)/(8/sqrt(150))
= 12.25
2. In the example above:
What is the null hypothesis?
Here the null hypothesis is
H0: The reading incentive program was not successful in increasing the number of books
read by 3rd graders,
that is m ≤ 47
What is the alternate hypothesis?
Here the alternative hypothesis is
Ha: The reading incentive program was successful in increasing the number of books
read by 3rd graders, that is
m > 47
Write these hypotheses in words and in symbols.
Here we want to test the null hypothesis
H0: The reading incentive program was not successful in increasing the number of books
read by 3rd graders
Against the alternative hypothesis
Ha: The reading incentive program was successful in increasing the number of books
read by 3rd graders.
That is we want to test H0:m ≤ 47 against Ha: m > 47
Is this a one- or two-tailed test?
This is a one-tailed (Upper-tailed) z- test for a population mean
3. In the example above, use your calculation of the test statistic and the table in the back
of your book to provide the best estimate for p (probability).
Here it is given that, m = 47, σ = 8, x-bar = 55, n = 150
The test statistic used is
z = (x-bar - m)/( σ /sqrt(n))
= (55 - 47)/(8/sqrt(150))
= 12.25
p- value for z = 12.25 is 0.0000
Decision Rule: Reject H0 if the p- value for the sample < alpha
What does p represent?
The p-value is the probability, if H0 is actually true, that random chance could give us the
result we got.
What does alpha represent?
Alpha represents the probability of rejecting a true null hypothesis
In order to support our alternate hypothesis, what should the relationship be between
alpha and our p value?
p-value < alpha
4. In the example above, if alpha = .05, have we found support for our alternate
hypothesis? Do we find support at alpha = .10? .01?
In all three cases, we find support for our alternate hypothesis.
5.In testing the hypothesis, H0: ? ? 28.7 and Ha: ? < 28.7, using the p-value approach, a
p-value of 0.0764 was obtained. If ? = 9.8, find the sample mean which produced this pvalue given that the sample of size n = 40 was randomly selected. (Hint: you are working
backwards on this problem. Using the p-value, determine the value of z. Put it into the z
equation to calculate the sample mean.)
This is a lower-tailed z- test and for p = 0.0764, z = -1.43
z = -1.43 = (28.7 - m)/(9.8/sqrt 40) = (28.7 - m)/1.5495
This gives m = 30.92
6. To test the null hypothesis that the average lifetime for a particular brand of bulb is 750
hours versus the alternative that the average lifetime is different from 750 hours, a sample
of 75 bulbs is used. If the standard deviation is 50 hours and ? is equal to 0.01, what
values for will result in rejection of the null hypothesis. Remember to determine whether
this is a one- or two-tailed test. (Hint: This is similar to problem #4 but it is NOT a onetailed test!)
This is a two-tailed z- test for a population mean
We reject the null hypothesis if the p- value for the sample of bulbs < 0.01
z- score for a = 0.01 is 2.576
2.576 = (x-bar - 750)/(50/sqrt 75)
x-bar (Lower limit) = 750 - 2.576(50/sqrt 75) = 735.13 hours
x-bar (Upper limit) = 750 + 2.576(50/sqrt 75) = 764.87 hours
The null will be rejected if the mean life is < 735.13 hours or > 764.87 hours
7. Describe the action that would result in a type I error and a type II error if each of the
following null hypotheses were tested:
H0: There is no waste in US Defense Department spending
H0: This fast-food menu is not low fat
a) We commit Type I error if there is no waste in US Defense Department Spending but
our test rejects the H0
We commit Type II error if there is waste in US Defense Department Spending but our
test accepts the H0
(b) We commit Type I error if the fast-food menu is not low fat but our test rejects the H0
We commit Type II error if there is low fat but our test accepts the H0
8. For the following z values of a 2-tailed test at alpha = .05, use your table and indicate
whether they fall within the critical region. If they do, what can we say about our null
hypothesis? Our alternate hypothesis?
z = 1.60
Here the critical value is 1.96. Since |z| = 1.60 <1.96 , we fail to reject H0
z = - 1.25
Here the critical value is 1.96. Since |z| = 1.25 <1.96 , we fail to reject H0
z = - 2.45
Here the critical value is 1.96. Since |z| = 2.45 >1.96 , we reject H0 and accept Ha
z = 1.98
Here the critical value is 1.96. Since |z| = 1.98 >1.96 , we reject H0 and accept Ha
9. Imagine our calculated p for a 1-tailed test is .047. If we have set alpha at .05, what can
we say? What could we say if this was a 2-tailed test? Which test has more power, i.e. the
ability to detect differences that actually exist
In case of one-tailed test, since 0.047 < 0.05, we would reject H0 and accept Ha.
(But this is a critical case since 0.047 is quite close to 0.05)
In case of two-tailed test, the rejection region narrows down and this makes accepting the
null hypothesis more probable.
10. What is the critical value of z that separates the lower 95% of the standard normal
distribution from the upper 5%? (Points: 4)
1.282
1.645
1.960
2.333
Answer : 1.645
11. What is the critical value of z that separates the lower 5% of the standard normal
distribution from the upper 95%? (Points: 4)
-1.282
-1.645
-1.960
-2.333
Answer: -1.645
12. What are the critical values of z that separate the middle 95% of the standard normal
distribution from the upper and lower 2.5%? (Points: 4)
+/- 1.282
+/- 1.645
+/- 1.960
+/- 2.333
Answer: +/- 1.960