1. What is pooled variance and why is it important?
Suppose we have two samples from two populations. Also we know that
both the populations have the same variance 𝜎 2 .
Then we will be having two sample variances 𝑠1 2 and 𝑠1 2 . They are defined
by
𝑠1 2 =
∑(𝑥1 −𝑥̅ 1 )2
𝑛1 −1
and 𝑠2 2 =
∑(𝑥2 −𝑥̅2 )2
𝑛2 −1
Both are estimates of the same population varinace 𝜎 2 . So we want to
combine these two estimates to a pooled estimate. It is defined as
(𝑛1 −1)𝑠1 2 +(𝑛2 − 1)𝑠2 2
𝑠 =
𝑛1 + 𝑛2 − 2
In hypothesis testing, when the common population variance is unknown,
we make use of the pooled sample variance.
2
2. Explain what interval data is and give an example:
Interval data is continuous data where differences are interpretable, but
where there is no "natural" zero. A good example is temperature in Fahrenheit
degrees.
Ratios are meaningless for interval data. You cannot say, for example, that one day
is twice as hot as another day.
4. Write the formula for pooled variance.
It is defined as
(𝑛1 −1)𝑠1 2 +(𝑛2 − 1)𝑠2 2
𝑠 =
𝑛1 + 𝑛2 − 2
2
𝑛1 and 𝑛2 are sizes of the two samples𝑠1 2 and 𝑠2 2 are the variances of the
two samples.
5. Please analyze the following data using the T-test for Unpaired data: A
school district wants to determine if the girls are equal to boys in test
scores. They took a random sample of 22 8th grade girls and 24 8th grade
boys. The district looked at recent CAT scores and found the mean score
for girls to be xbar1 = 25 and the mean score for boys to be xbar2 = 26.
The standard deviation for the girls is s1 = 2.2 and for the boys the
standard deviation is s2 = 3.4. Is the school district correct in assuming the
girls are equal in performance to boys. They are 95% sure their assumption
is correct.
n1 = 22
n2 = 24
xbar1 = 25
xbar2 = 26
s1 = 2.2
s2 = 3.4
pooled variance = 𝑠 = √
(𝑛1 −1)𝑠1 2 +(𝑛2 −1)𝑠2 2
𝑛1 +𝑛2 −2
=√
(22−1)(2.22 )+(24−1)(3.4 2 )
22+24−2
101.64 + 265.88
=√
= √8.3527 = 2.8901
44
Null hypothesis H0: μ1 = μ2
The mean scores of girls and boys are equal.
Alternative hypothesis H1: μ1 ≠ μ2
The mean scores of girls and boys are different.
The test statistic is
𝑥̅1 − 𝑥̅2
25 − 26
𝑇=
=
= −1.1723
1
1
1
1
𝑠√𝑛 + 𝑛
2.8901√ +
22 24
1
2
Degrees of freedom = 𝑛1 + 𝑛2 − 2 = 22 + 24 − 2 = 44
5% Critical value of t with 44 degrees of freedom = 2.0154
Critical region is |𝑇| > 2.0154
The computed value T is not in the critical region.
So the does not provide sufficient evidence to reject the null hypothesis.
So we conclude that the school district is correct in assuming the girls are
equal in performance to boys.
6. Please analyze the following data using the Z-test for Unpaired data A
candy company wants to identify whether or not the size of its' candy bars
are the same length from one day to another. On day 1 they sample 52
candy bars with an average mean of 5.4 inches and on day 2 they sample
52 bars again with and average mean of 5.6 inches. Both days have
sample standard deviations of 3.4. Is the average mean of size different
from day 1 to day 2?
𝑥̅1 = 5.4, 𝑥̅2 = 5.6, 𝑛1 = 52, 𝑛2 = 52, 𝜎 = 3.4
Null hypothesis 𝐻0 : 𝜇1 = 𝜇2
The means of day1 and day 2 are the same.
Alternative hypothesis 𝐻1 : 𝜇1 ≠ 𝜇2
The means of the two days are different.
The test statistic is
𝑥̅1 − 𝑥̅2
5.4 − 5.6
𝑍=
=
= −1.0198
1
1
1
1
𝜎√𝑛 + 𝑛
3.4√ +
52 52
1
2
The critical region is |𝑍| > 1.96
Decision
The computed value of Z is not significant.
So the sample does not provide sufficient evidence to reject the null
hypothesis. So we conclude that the average mean size is the same from
day to day.
7. A sample of 40 observations is selected from one population. The
sample mean is 102 and the sample standard deviation is 5. A sample of
50 observations is selected from a second population. The sample mean is
99 and the sample standard deviation is 6. Conduct the following test using
95% level of confidence. Z-test for Unpaired data A. Is it a 1 or 2 tail test B.
Compute the statistical calculation C. What is your decision regarding H1
𝑛1 = 40
𝑥
̅̅̅1 = 102
𝑠1 = 5
𝑛2 = 50
𝑥̅2 = 99
𝑠2 = 6
𝐻0 : 𝜇1 = 𝜇2
𝐻1 : 𝜇1 ≠ 𝜇2
𝑥̅1 − 𝑥̅2
1
102 − 99
𝑧=
=
1
1
2
2√
2
2
+
(𝑛
−1)𝑠
+(𝑛
−
1)𝑠
1
1
2
2
𝑛1 𝑛2 √39(5 ) + 49(6 ) √ 1 + 1
√
𝑛1 + 𝑛2 − 2
40 50
40 + 50 − 2
= 2.5349 > 1.96
It is a two tail test.
Reject
𝐻0 : 𝜇1 = 𝜇2
So we accept the alternative hypothesis H1. The means are different.
8. A sample of 65 observations is selected from one population. The
sample mean is 2.67 and the sample standard deviation is .75. A sample of
50 observations is selected from a second population. The sample mean is
2.59 and the sample standard deviation is .66. Conduct the following test
using 95% level of confidence. Z-test for Unpaired data a. Is it a 1 or 2 tail
test b. Compute the statistical calculation c. What is your decision
regarding H1 d. What is the P-value 9. The Gibbs Baby Food Company
wishes to compare the weight gain of infants using their brand versus their
competitor's brand. A sample of 40 babies using the Gibbs products
revealed a mean weight gain of 7.6 pounds in the first 3 months after birth.
The standard deviation of this sample is 2.3 pounds. A sample of 55 babies
using the competitors brand revealed a mean increase of 8.8 pounds with a
standard deviation of 2.9 pounds. At 95% level of confidence, can we
conclude that the babies using the Gibbs product gained LESS weight? Ztest for Unpaired data 10. What key words tell you that the Hypothesis is to
be set up as a 1 Tail UPPER?