1. 10.6 Tests Concerning Differences Between Means for Large Samples
Advanced Math Topics
10.6 Tests Concerning Differences Between Means for Large Samples

2. A little refresher of the previous sections in this chapter… tests concerning
means for 1) large samples and 2) small samples…

3. Alternative Hypothesis, H1: μ < 75
The average score of all 6th graders in a school district on a standardized
math test is 75 with a standard deviation of A random sample of 100
students at a certain school was taken, and their mean score was 71. Does this
indicate that the students of this school are significantly less skilled in math
than the average student in the district? Use a 1% level of significance.
Null Hypothesis, H0: μ = 75
Alternative Hypothesis, H1: μ < 75
If the sample mean is within our significance level, we will accept the null hypothesis. If the sample mean is outside our
significance level, we will reject the null hypothesis and accept the alternate hypothesis.
0.49
0.01
μ = 75
z = -2.33
Since the alternative hypothesis is that μ < 75, this claim
can be shown if the sample mean is much
less than 75. Thus, it is called a one-tailed test.
z = -4.94
If the results of the sample have a z-score less than
-2.33, then we can reject the null hypothesis.
Since the z-score is in the rejection region, our decision is that we must
reject the null hypothesis. The students at this school are significantly
less skilled than the average student in the district.

4. manufacturer’s claim? Use a 5% level of significance.
A manufacturer claims that each bag of mixed nuts sold contains an average of
10 cashew nuts. A sample of 15 cans has an average of 8 cashew nuts with
a standard deviation of 3. Does this indicate that we should reject the
manufacturer’s claim? Use a 5% level of significance.
Null Hypothesis, H0: μ = 10
Alternative Hypothesis, H1: μ ≠ 10
It does not state that we are looking for strictly MORE THAN or LESS THAN 10 cashews,
and we are only looking to reject the claim that there is an average of 10 cashews,
therefore this is a two-tail test and we use ≠.
0.025
0.025
μ = 10
t =
z = 2.145
Since the sample is small, we use the t-table in the
back. The degrees of freedom is = 14. You can
either draw a picture and label each rejection region
or divide the significance level by 2 (because
it is a two-tail test) to find the 0.05/2 = t0.025.
t = -2.58
Since the t-score is in the rejection region, our decision is that must reject
the claim that each bag has an average of 10 cashews.

5. an IQ test to 50 men and 50 women. The men had an average score of
In this lesson, similar strategies can be applied.
You will be given two sample means to compare. Example: A professor gave
an IQ test to 50 men and 50 women. The men had an average score of
78 with a standard deviation of 7 and the women had an average score of 81
with a standard deviation 9. It is important to decide if the observed
difference between the means is purely by chance or if the difference is
significant.
To do so we need six variables to represent the six given numbers.
x1= the first sample mean
x2= the second sample mean
s1= the first standard deviation
s2= the second standard deviation
n1= the first sample size
n2= the second sample size
The formula…
Tails What the book will say…
A claim that one mean is
higher or lower.
Is there any significant
difference?
Compare this number to the z-score (in the back of the book or
you may have memorized) for the significance level and if it is
a one-tail or two-tail test.

6. recent male college graduate. A survey of 30 women found their average
A sociologist claims that a recent female college graduate earns less than a
recent male college graduate. A survey of 30 women found their average
salary to be $29,000 with a standard deviation of $600. A survey of 40 men
found their average salary to be $29,700 with a standard deviation of $900.
Do these figures support the claim that women earn less? Use a 1% level of
significance.
One tail with 1% significance.
0.49
0.01
z = -2.33
There is a significant difference between
the starting salaries of men and women.
z = -3.90

7. z = 2.62 which is outside the acceptance region of z = 1.96.
From the HW P. 505
3) Forty-five traffic signals that were painted with one brand of paint lasted an average
of 3.4 years before being repainted with a standard deviation of 0.69 years. Sixty-two
traffic signals painted with a second brand of paint lasted an average of 2.98 years with
a standard deviation of 0.97 years. Is there a significant difference between the average
life of the two brands of paint? Use a 5% level of significance.
z = which is outside the acceptance region of z = 1.96.
There is a significant difference in the means.

8. HW
P. 505 #3-10