1. Hypothesis Testing – Two Means(Small, Independent Samples)
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Section 11.2
Hypothesis Testing – Two Means(Small, Independent Samples)

2. The samples are independent.
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Criteria for hypothesis testing with small samples:
The samples are independent.
Each population from which the samples were drawn has a normal distribution.
n < 30 for one or both samples.
 is unknown for both populations.

3. If the samples can be pooled when variances are equal:
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Test Statistic for Small Samples, n < 30:
If the samples can be pooled when variances are equal:
with d.f. = n1 + n2 – 2
Formula for pooled standard deviations:
Formula for the value of t:

4. If the samples cannot be pooled when variances are not equal:
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Test Statistic for Small Samples, n < 30:
If the samples cannot be pooled when variances are not equal:
where the degrees of freedom is the smaller of n1 – 1 and n2 – 1

5. The critical values for n < 30 are found from the t-distribution.
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
To determine if the test statistic calculated from the sample is statistically significant we will need to look at the critical value.
The critical values for n < 30 are found from the
t-distribution.

6. For a left-tailed test, reject H0 if t ≤ –t
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Rejection Regions:
For a left-tailed test, reject H0 if t ≤ –t
For a right-tailed test, reject H0 if t ≥ t
For a two-tailed test, reject H0 if | t | ≥ t/2

7. State the null and alternative hypotheses.
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Steps for Hypothesis Testing:
State the null and alternative hypotheses.
Set up the hypothesis test by choosing the test statistic and determining the values of the test statistic that would lead to rejecting the null hypothesis.
Gather data and calculate the necessary sample statistics.
Draw a conclusion.

8. First state the hypotheses: H0: Ha: m1 – m2 ≥ 0 m1 – m2 < 0
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Draw a conclusion:
A home improvement warehouse claims that people interested in home improvement projects can save time by attending their workshops. Specifically, the warehouse claims that people who attend their tiling workshops take less time to complete comparable projects than those who do not. To test the claim, a group of 10 people who attended their workshops is later surveyed about the time it took to finish the project. This group spent an average of 14.1 hours to complete their projects with a standard deviation of 2.3 hours. Another 10 people chosen for the study did not attend the workshops, and they spent on average 15.0 hours with a standard deviation of 2.4 hours. Test the claim at the 0.01 level of significance. Assume that the populations are normally distributed and that the variances of the populations are the same.
Solution:
First state the hypotheses:
H0:
Ha:
m1 – m2 ≥ 0
m1 – m2 < 0

9. Next, set up the hypothesis test and state the level of significance:
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11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Solution (continued):
Next, set up the hypothesis test and state the level of significance:
a = 0.01, d.f. = n1 + n2 – 2 = – 2 = 18
t0.01 =
Reject if t ≤ –t , or if t ≤ –
Gather the data and calculate the necessary sample statistics:
n1 = 10, 1 = 14.1, s1 = 2.3, n2 = 10, 2 = 15.0, s2 = 2.4
–2.552
 –0.856

10. Finally, draw a conclusion:
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Solution (continued):
Finally, draw a conclusion:
Since t= is greater than t = we will fail to reject the null hypothesis. There is not sufficient evidence at the 0.01 level of significance to say that people who complete the tiling workshops take less time to complete their projects than those who do not.
–0.856

11. First state the hypotheses: H0: Ha:
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Draw a conclusion:
The Smith CPA Firm claims that their clients obtain larger tax refunds than clients of their competitor, Jones and Company CPA. To test the claim, 15 clients from the Smith firm are selected and found to have an average tax refund of $942 with a standard deviation of $103. At Jones and Company, 18 clients are surveyed and found to have an average refund of $898 with a standard deviation of $95. Test the claim at the 0.05 level of significance. Assume that both populations are normally distributed. Because the two firms are located in difference parts of a large city, assume the variances for the two populations are not equal.
Solution:
First state the hypotheses:
H0:
Ha:
Next, state the level of significance:
= 0.05, d.f. = d.f. is the smaller of n1 – 1 and n2 – 1
t0.05, 14 =
Reject if t ≥ t , or if t ≥
m1 – m2 ≤ 0
m1 – m2 > 0
1.761

12. Gather the data and calculate the necessary sample statistics:
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Hypothesis Testing (Two or More Populations)
11.2 Hypothesis Testing – Two Means (Small, Independent Samples)
Solution (continued):
Gather the data and calculate the necessary sample statistics:
n1 = 15, 1 = 942, s1 = 103, n2 = 18, 2 = 898, s2 = 95
1.266
Finally, draw a conclusion:
Since t=1.266 is less than t (1.761), we will fail to reject the null hypothesis. There is not sufficient evidence at the 0.05 level of significance to support Smith CPA’s claim that their clients receive larger tax refunds than clients of Jones and Company.
1.266

13. It resides in the left tail Thus, Critical-T = -2.0860
μ1 - μ2 = 0 Population Means are the same
μ1 - μ2 ≠ 0 Population Means are the different
Critical t =

15. =
Std Error =
SQRT[ x ( )] =

16. Confidence Interval = (m1 – m2 ) ± (Critical-T x Std Error)
Confidence Interval = (m1 – m2 ) ± (Margin of Error)
Lower = ( ) - (2.086 x ) =
Upper = ( ) + (2.086 x ) =

19. = 17.52 =
= (77.5 – 84.3) ÷ =

20. Critical-T (0.01,24) = 2.492
It resides in the left tail
Thus, Critical-T =

22. μ1 ≥ μ2 Population Mean 1 Greater than Population Mean 2
μ1 < μ2 Population Mean 1 Less than Population Mean 2
Reject if t < -t,d.f. or p-value < 
Reject if |t| > |2.492| or t <
Computed-t is and it is < , reject the null hypothesis
p-value is and it is less than 0.01, reject the null hypothesis
Reject the Null Hypothesis. There is enough evidence that μ1 is lower than μ2