1. Testing Difference among Mean, Variances, and Proportions. Chapter 10

2. Introduction
The basic concepts of hypothesis testing were just discussed in Chapter 9. The tests we used compared sample means, variances, or proportions to a specific mean, variance, or proportion.
There are however many times when researchers wish to compare two means, two variances, or two proportions using experimental and control groups.
In this chapter we will once again be using those tests to reject or keep null hypothesis.

3. Testing the Difference between 2 Means (Large Samples)
The theory behind testing the difference between 2 means is based on selecting pairs of samples and comparing the means of the pairs.
All possible pairs of samples are taken from the population. The means are computed and then subtracted and the differences are plotted. Most difference turn out to be zero but occasionally there are larger differences due to chance.
If differences are plotted the curve will be shaped like the normal distribution curve.

4. Assumptions for the Test to Determine Difference between 2 Means
#1 The samples must be independent of each other. There can be no relationship between the subjects in each sample.
#2 The populations from which the samples were obtained are normally distributed, and the standard deviations of the variables must be known, or the samples must be greater than or equal to 30.

6. Testing the Difference between 2 Means (Large Samples)
The null and alternative hypotheses are written differently.
Examples: H₀: 𝜇₁ = 𝜇₂ and H₁: 𝜇₁ ≠𝜇₂
H₀: 𝜇₁ ≤ 𝜇₂ and H₁: 𝜇₁ >𝜇₂
H₀: 𝜇₁ ≥ 𝜇₂ and H₁: 𝜇₁ <𝜇₂
The critical values will be found in the same way.
If 𝜎 2 1 and 𝜎 2 2 are not known researchers can use the variances obtained from each sample but both sample sizes MUST be 30 or more. We can replace 𝜎 2 with 𝑠 2 .

7. Testing 2 Means Example #1
A survey found that the average hotel rate in New Orleans is $88.42 and avg. in Phoenix is $ Assume samples were from 50 hotels each. Standard deviations are $5.62 and $4.83 respectively. At ∝ =0.05, can it be concluded that there is no significant difference in the rates?
Step 1: State Hypotheses
Step 2: Determine Critical Values
Step 3: Determine Test Value
Summarize

8. Testing 2 Means Example #2
Company 1
Company 2
X1 = 16 hours
X2 = 18 hours
𝜎1=3.2
𝜎2=3
n1 = 30
n2 = 30
A company wishes to test claim that there is no difference between 2 overnight delivery companies in the speed with which their material is delivered. The avg. speed over a 30-day period is shown above. At ∝ =0.01, is there enough evidence to support the claim that there is no difference between the delivery times of the 2 companies?

9. Testing 2 Means Example #3
York School District
Somerville School District
X1 = 650
X2 = 562
𝜎1=99
𝜎2=99
n1 = 30
n2 = 30
A college admissions officer believes that the students enrolling from York School District have higher SAT scores than those from Somerville School District. The results from the sampling of 30 students is shown above. At ∝ =0.05, is there enough evidence to support the claims of the admissions officer?

10. Testing 2 Means (Your Turn)
Spouses
Significant Others
X1 = 2
X2 = 1.7
𝝈𝟏=𝟎.𝟔
n1 = 120
𝝈𝟐=𝟎.𝟕
n2 = 34
A study was conducted to see if there was no difference between spouses and significant others in coping skills when living with in-laws. These skills were measured by questionnaire responses. The results of the 2 groups are given above on one factor, ambivalence. At ∝ =0.10, is there no difference in the means of the 2 groups?

11. 𝝁𝟏 − 𝝁𝟐 ≠𝟎
Sometimes researchers are interested in testing specific difference in means other than zero. Example: They might hypothesize that nursing students at a community college are, on the average 3.2 years older than those at large university. In this case hypotheses are
H0 𝜇1 −𝑢2≤3.2
H1 𝜇1 −𝑢2>3.2 Claim
Z test formula would remain the same but the value of 𝜇1 −𝑢2 = 3.2

12. Confidence Intervals for 2 Means
Confidence intervals of the difference between 2 means can also be found.
Formula:
Z ∝/2 is found in the same way as in Ch. 8.
90% = 1.65, 95% = 1.96 and 99% = 2.58

13. Confidence Interval Example #1
Mathematics Majors
Computer Science Majors
X1 = 83.6
X2 = 79.2
𝜎1=4.3
𝜎2=3.8
n1 = 36
n2 = 36
Two groups of students are given a problem-solving test, and the results are compared. Find the 90% confidence interval of the true difference in means.

14. Confidence Interval (Your turn)
Brand X
Brand Y
X1 = 9.2 volts
X2 = 8.8 volts
𝜎1=0.3 𝑣𝑜𝑙𝑡𝑠
𝜎2=0.1 𝑣𝑜𝑙𝑡𝑠
n1 =27
n2 = 30
The data above shows two brands of batteries that are tested and their voltage compared. Find the 95% confidence interval of the true difference in the means.

15. Testing the Difference between Two Variances
In addition to comparing two means, statisticians are also interested in comparing two variance or standard deviations.
For the comparison of these an F-Test is used.
This test should not be confused with the chi-square test which only compared single sample variances to a specific population variance.

16. Characteristics of the F Distribution
The values of F cannot be negative, because variances are always positive or zero.
The distribution is positively skewed.
The mean value of F is approximately equal to 1.
The F distribution is a family of curves based on degree of freedom of the variance of the numerator (d.f.N) and the degree of freedom of the variance of the denominator (d.f.D)

17. Formula for the F Test
The numerator MUST be the larger of the 2 variances.
The F test has 2 terms for the degrees of freedom: that of the numerator n1 – 1 and that of the denominator n2 – 1, where n1 is the sample size from which the larger variance was obtained.

18. C.V. for F Test We will be using Table H in appendix C to find C.V.
There are separate tables for different ∝.
If a two tailed is being conducted we must divide the ∝ by 2 and use the correct table.
When the degrees of freedom value cannot be found in the table the closest value on the smaller side should be used. Example if d.f.N = 14, this falls between 12 and 15 on chart therefore 12 should be used to be on the safe side.

19. C.V. Examples for F Test
Example #1: Find C.V. for right-tailed test where ∝ =0.05 and d.f.N = 15 and d.f.D = 21.
Example #2: Find C.V. for 2-tailed test where ∝ =0.05 and the sample size for variance of numerator was 21 and the sample size from variance of the denominator was 12.
Example #3: Find C.V. for 2-tailed test where ∝ =0.10 and the sample size for variance of numerator was 13 and the sample size from variance of the denominator was 20.

20. Testing Difference between 2 Variances Example #1
When writing hypotheses the 𝜎 2 replaces the 𝜇 symbols.
If standard deviations are given instead of variances then they must be squared for the formula.
Once again we will use the 5 steps.
Example #1: A medical researcher wishes to see if the variances of heart rates of smokers are different from the variances of heart rates of non-smokers. 2 samples are selected. Smokers n1 = 26 and s² = 36. Non-smokers n2 = 18 and s² = 10. At ∝ =0.05 , is there enough evidence to support the claim?

21. Testing Difference between 2 Variances Example #2
An instructor hypothesizes that the standard deviation of the final exam grades in her class is larger for males than females. The data from the final exam is shown below. Is there enough evidence to support her claim at ∝ =0.01?
Males n1 = 16, s1 = 4.2
Females n2 = 18, s2 = 2.3

22. Testing Difference between 2 Variances Examples #3 and 4 (All yours)
In a hospital study it was found that the standard deviations of the sound levels from 20 areas designated as “casualty doors” was 4.1 dBA and the standard deviation of 24 areas designated as operating theaters was 7.5 dBA. At ∝ =0.05, is there enough evidence to support the claim that there is no difference in the standard deviations?
Researcher claims that the variance of IQ scores for women in psychology is larger than IQ scores for men. A sample of 22 IQ scores for women had variance of 192 and for 18 men variance was 84. At ∝ =0.05, is there enough evidence to support the claim?

23. Differences between 2 Mean: Small Independent Samples
Used when the population standard deviations are not known, and one or both of the sample sizes in less than 30. In these cases the t-test will be used to test the differences between means.
There are actually 2 different options for the use of the t-test.
#1: Used when the variances of the populations are not equal.
#2: Used when the variances are equal.

24. Formulas for the T test- Testing Difference between Means- Small Sample
The C.V. will be determined by using the T table. The d.f. are equal to the smaller of n1 – 1 or n2 – 1.
Variances are assumed to be unequal.
Going to be using this formula exclusively.

25. Formulas for the T test- Testing Difference between Means- Small Sample #2
Where the d.f. = n1 + n2 – 2
The variances are assumed to be equal.

26. Steps to Solving Difference of 2 Means Small Indy Samples
Step 1: State hypothesis and identify claim.
Step 2: Find the C.V. by using the T table.
The d.f. = n1 – 1 or n2 – 1, whichever is smaller.
Step 3: Determine the T.V. by using formula
Step 4: Summarize answer

27. T test – Difference between 2 Means- Example #1 (Small samples)
Male
Female
X1 = $23,800
X2 = $23,750
S1 = $300
S2 = $250
n1 = 16
n2 = 20
A researcher suggests that the male nurses earn more then female nurses. The data above shows the results of the survey. Is there enough evidence to support the claim that males earn more than females? Let ∝ =0.05.

28. T test – Difference between 2 Means- Example #2 (Small samples)
The average monthly premium paid by 12 administrators for hospitalization insurance is $56 with a standard deviation of $3. The avg. monthly premium paid by 27 nurses is $63, with a standard deviation of $ At ∝ =0.05, is there enough evidence to say that the nurses pay more than the administrators?

29. T test – Difference between 2 Means- Example #3 (Small samples)
The local branch of the IRS spent an average of 21 minutes helping each of 10 people prepare their taxes, with a standard deviation of A volunteer tax preparer spent an average of 27 minutes helping 14 people, with a standard deviation of 4.3 minutes. At ∝ =0.02, is there a difference in the average times spent by the two services?

30. Confidence Intervals for Difference of 2 Means (Small Samples)
Basically same for intervals of larger samples except using a 𝑡 ∝/2 value.

31. Confidence Interval Example #1 (Small Samples)
Average size of a farm in Greene County, PA is 199 acres. Avg. size in Indiana County is 191 acres. Deviations of the 2 samples was 12 acres and 38 acres, respectively. Sample sizes were 10 farms and 8 farms respectively. Let ∝ =0.05, find the 95% confidence interval for the difference of the 2 means. Is there a significant difference between the acreage?

32. Confidence Interval Example #2 (Small Samples)
Sample of 15 teachers rom Rhode Island has avg. salary of $35,270 with a standard deviation of $3,256. A sample of 30 teachers from N.Y. has an avg. salary of $29,512 with a standard deviation of $1,432. Using ∝ =0.02, find the 99% CI for the difference of two means. Is there a significant difference in the salaries between the states?

33. Testing the difference between 2 Means: Small Dependent Samples
Previously discussed comparing 2 sample means when samples were independent. When dependent a different version of the t test is used.
Samples are considered dependent when the subjects are paired or matched in some way.
Example: Researcher wants to design an SAT prep course to help students raise their scores. They are first given the test and then re-tested after the course. The difference between the 2 tests is compared.
The new test employs the difference in values of the matched pairs.

34. Testing the Difference between 2 means (Small Dependent Samples)
Step 1: Hypotheses will be given the new symbol of 𝜇 𝐷 .
Step 2: Find the Critical Value.
Step 3: Find the test value by completing the following.
#1: Find the differences between the pairs of data. 𝐷= 𝑋 𝑋 2
#2: Find the mean of the differences. ⅀𝐷 𝑛 where n = number of data pairs.( D )
#3: Find the standard deviation of the differences. SD= ⅀ 𝐷 2 − (⅀𝐷)² 𝑛 / 𝑛 −1
#4: Finally find the test value ( D - 𝜇 𝐷 ) / ( 𝑆 𝐷 ÷ 𝑛 )
Step 4: Summarize

35. Example 1 Testing Difference between 2 Means (Small Dependent)
Subject 1 2 3 4 5 6
Before (X1)
210
235
208
190
172
244
After (X2)
170
188
173
228
A dietitian wishes to see if a person’s cholesterol level will change if the diet is supplemented by a certain mineral. 6subjects were pretested and then took the mineral supplement for 6 weeks. The results are shown above. Can it be concluded that the cholesterol levels have been changed at ∝ =0.10?
Step 1: Find the difference between the means and find that sum.
Step 2: Find the average of the differences.
Step 3: Square the differences and find that sum.
Step 4: Find the standard deviation. SD= ⅀ 𝐷 2 − (⅀𝐷)² 𝑛 / 𝑛 −1
Step 5: Find the test value. ( D - 𝜇 𝐷 ) / ( 𝑆 𝐷 ÷ 𝑛 )
Summarize