11--Inferences Chapter on Two Samples
11.1--Inference about Two Means: Dependent Samples (or Matched-Pairs) t hypothesis test t confidence interval 11.2--Inference about Two Means: Independent Samples t hypothesis test--unequal population variances t confidence interval t hypothesis test--equal population variances 11.3--Inference about Two Population Proportions Z hypothesis test Z confidence interval Sample size Example--Comparison of two teaching methods for statistics: Traditional method involving class lectures Computer method using web-based instruction Example--Comparison of pig gains on two diets.
Independent versus Dependent Sampling Independent sampling--when the individuals selected for one sample do not dictate which individuals are to be in a second sample. Dependent sampling--when the individuals in two samples are somehow related: husband-wife siblings similar characteristics same person same year
Example 1, p. 508, provides examples of independent versus dependent samples.
Matched-Pairs Design using the t-Distribution--Hypothesis Test Regarding the Difference of Two Means (d) with d Unknown Assumptions: The sample is obtained using simple random sampling; The sample data are matched-pairs; The differences are normally distributed or the sample size, n, is "large" (n>30). Step 1: A claim is made regarding the mean difference from matched-pairs data (d). The null and alternative hypotheses can be structured in three ways:
Two-Tailed H0: d = 0 H1: d 0 d<0 or d>0 Left-Tailed H0: d = 0 H1: d < 0 Right-Tailed H0: d = 0 H1: d > 0
Step 2: Select a level of significance, , which is generally chosen to be 0.10, 0.05, or 0.01. The significance level is used to determine the critical value. Critical value is the t-value that separates the rejection and nonrejection regions. The rejection region (or critical region) is the set of all values of the test statistic (defined in step 3) such that the null hypothesis is rejected.
Step 3: Calculate the test statistic or calculated t-value.
Note: Statistical inference methods on matched-pairs data use the d - d d - 0 d t= = = , same methods as inference on a single population mean with sd sd sd unknown, except that differences are analyzed. n n n which follows Student's t-distribution with df=n-1. The values of d and sd are the mean and standard deviation of the differenced data. The test statistic (t) measures the number of standard deviations that the sample mean, d , is from the assumed population mean, d=0.
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Step 4: Draw a conclusion:
Compare the calculated t-value (or test statistic) to the critical t-value and state whether or not H0 is rejected at the specified .
Two-Tailed Left-Tailed Right-Tailed
If t < -t or t > t
2
2
If t < -t reject the null hypothesis.
If t > t reject the null hypothesis.
reject the null hypothesis
Interpret the conclusion in the context of the problem.
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Example 2, p. 510--Testing a Claim Regarding Matched-Pairs Data
Problem: Professor Andy Neill measured the time (in seconds) required to catch a falling meter stick for 12 randomly selected students' dominant hand and nondominant hand. Professor Neill claims that the reaction time in an individual's dominant hand is less than the reaction time in their nondominant hand. Test the claim at the =0.05 level of significance. Step 1: Null and Alternative Hypotheses: H0 : d = 0 d < 0 H1 : Step 2: Select = 0.05 and find the critical value of t (df=12-1=11).
Step 3: Draw a random sample of n=12 students and measure their reaction times in the dominant and nondominant hands. Calculate the sample mean, d , sample standard deviation, sd, and the test statistic, t. A 1 Student 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 14 15 16
t=
B C D E Dominant Nondominant Difference, Hand, X1 Hand, X2 di=X1-X2 (d i - d ) 0.177 0.179 -0.002 0.011167 0.210 0.202 0.008 0.021167 0.186 0.208 -0.022 -0.008833 0.189 0.184 0.005 0.018167 0.198 0.215 -0.017 -0.003833 0.194 0.193 0.001 0.014167 0.160 0.194 -0.034 -0.020833 0.163 0.160 0.003 0.016167 0.166 0.209 -0.043 -0.029833 0.152 0.164 -0.012 0.001167 0.190 0.210 -0.020 -0.006833 0.172 0.197 -0.025 -0.011833 Sum= -0.158 0.000000 -0.013167 =AVERAGE(D2:D13) d = s d= 0.01643 =STDEV(D2:D13)
sd =
F
(d i - d ) 2
0.000125 0.000448 0.000078 0.000330 0.000015 0.000201 0.000434 0.000261 0.000890 0.000001 0.000047 0.000140 0.002970
(d - 0) (-0.013167 - 0) - 0.013167 = = = -2.778 sd 0.01643 0.00474 12 n
(d i - d ) 2
n -1
__
=
0.00297 = 0.01643 12 - 1
Step 4: Conclusion--Because the calculated t = -2.778 is less than the critical t = -1.796 (and in the rejection region), reject H0 at the 0.05 significance level. There is sufficient evidence at the 0.05 significance level to support Professor Neill's claim that the mean reaction time in the dominant hand is less than the mean reaction time in the nondominant hand.
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Example 2, p. 510--Using Excel to Test a Claim Regarding Matched-Pairs Data
Problem (see previous page): This problem can be solved using Excel. One advantage is that
Excel provides a calculated P-value as well as a calculated t-value. The P-value serves as a measure of the strength of evidence against H0. A small P-value means that the null hypothesis is strongly rejected or the result is highly statistically significant.
Excel: Two-sample t-tests, Dependent Sampling (or Matched-Pairs): Step 1: Enter raw data in columns B and C (the Excel worksheet page is shown on the next page). Step 2: Select the Tools menu, highlight Data Analysis. Step 3: Select "t-test: Paired Two-Sample for Means." With the cursor in the "Variable 1 Range" box, highlight the data in column B. With the cursor in the "Variable 2 Range" box, highlight the data in column C. Enter the hypothesized mean difference (usually 0) and a value for alpha (e.g., =0.05). In the "Output Range" box, specify a cell for the output (upper left corner of the output range). Click OK.
From Tools menu select Data Analysis.
Select an Analysis Tool: t-Test: Paired Two Sample for Means
Highlight the data for Variable 1 and Variable 2
Enter the hypothesized mean difference of 0, and enter =0.05
Specify a cell for the output. 4
Output from Excel--t-Test: Paired Two Sample for Means
A B Dominant Hand, X1 0.177 0.210 0.186 0.189 0.198 0.194 0.160 0.163 0.166 0.152 0.190 0.172 C Nondominant Hand, X2 0.179 0.202 0.208 0.184 0.215 0.193 0.194 0.160 0.209 0.164 0.210 0.197 Sum= D Difference, di=X1-X2 -0.002 0.008 -0.022 0.005 -0.017 0.001 -0.034 0.003 -0.043 -0.012 -0.020 -0.025 -0.158 -0.013167 E F
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Student 1 2 3 4 5 6 7 8 9 10 11 12
(d i - d )
0.011167 0.021167 -0.008833 0.018167 -0.003833 0.014167 -0.020833 0.016167 -0.029833 0.001167 -0.006833 -0.011833 0.000000
(d i - d ) 2
0.000125 0.000448 0.000078 0.000330 0.000015 0.000201 0.000434 0.000261 0.000890 0.000001 0.000047 0.000140 0.002970
d =
t-Test: Paired Two Sample for Means Dominant Hand, X1 0.179750 Mean 0.000307 Variance 12 Observations 0.572 Pearson Correlation Hypothesized Mean 0 Difference 11 df -2.776 t Stat 0.00902 P(T<=t) one-tail t 1.796 Critical one-tail 0.018 P(T<=t) two-tail 2.201 t Critical two-tail
Nondominant Hand, X2 0.192917 0.000324 12
Confidence Interval for a Matched-Pairs Test:
(1-)100% CI: point estimate margin of error s d t d 2 n where t is computed using df=n-1.
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Independent-Samples Design using the t-Distribution--Hypothesis Test Regarding the Difference of Two Means with Unknown Population Standard Deviations (1 and 2) Assumptions: The samples are obtained using simple random sampling; The samples are independent; The populations from which the samples are drawn are normally distributed or the sample sizes are "large" (n1>30 and n2>30). Step 1: A claim is made regarding the two population means (1 and 2). The null and alternative hypotheses can be structured in three ways:
Two-Tailed H0: 1 = 2 H1: 1 2 1<2 or 1>2 Left-Tailed H0: 1 = 2 H1: 1 < 2 Right-Tailed H0: 1 = 2 H1: 1 > 2
Step 2: Select a level of significance, , which is generally chosen to be 0.10, 0.05, or 0.01. The critical value is determined using the smaller of n1-1 or n2-1 degrees of freedom. Critical value is the t-value that separates the rejection and nonrejection regions. The rejection region (or critical region) is the set of all values of the test statistic (defined in step 3) such that the null hypothesis is rejected.
Step 3: Calculate the test statistic or calculated t-value.
, 2 2 s1 s2 + n1 n2 which approximately follows Student's t-distribution. The test statistic (t) measures the number of standard deviations that the difference in sample means, ( x 1 - x 2 ), is from the assumed difference in the population means, (1-2)=0.
t=
( x1 - x 2 ) - ( 1 - 2 )
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Step 4: Draw a conclusion: Compare the calculated t-value (or test statistic) to the critical t-value and state whether or not H0 is rejected at the specified .
Two-Tailed Left-Tailed Right-Tailed
If t < -t or t > t
2
2
If t < -t reject the null hypothesis.
If t > t reject the null hypothesis.
reject the null hypothesis
Interpret the conclusion in the context of the problem. 6
Independent-Samples Design using the t-Distribution--Hypothesis Test Regarding the Difference of Two Means with Unknown Population Standard Deviations (1 and 2) Problem (p. 523): In the Spacelab Life Sciences 2 payload, 14 male rats were sent to space. Upon their return, the red blood cell mass (in milliliters) of the rats was determined. A control group of 14 male rats was held under the same conditions (except for space flight) as the space rats and their red blood cell mass was also determined when the space rats returned. Test the claim that the flight rats have a different mean red blood cell mass from the control rats at the = 0.05 significance level. Step 1: Null and Alternative Hypotheses:
H0: H1 :
1 = 2 or 1 - 2 = 0 1 2 or 1 - 2 0 1 < 2 or 1 > 2
Step 2: Select = 0.05 and find the critical value of t (df=14-1=13).
Step 3: Draw a random sample of n=28 rats and randomly assign them to the flight group or the control group. Calculate the sample means, x1 and x 2 , standard deviations, s1 and s2, and test statistic, t.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 A Flight, X1 8.59 8.64 7.43 7.21 6.87 7.89 9.79 6.85 7.00 8.80 9.30 8.03 6.39 7.54 B Control, X2 8.65 6.99 8.40 9.66 7.62 7.44 8.55 8.70 7.33 8.58 9.88 9.94 7.14 9.14
t=
( x 1 - x 2 ) - ( 1 - 2 ) s1 s + 2 n1 n2
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2
2
=
(7.881 - 8.430) - 0 1.017 2 1.005 2 + 14 14
=
- 0.549 = -1.437 0.3821288115
s1 =
( x1i - x 1 ) 2
n1 - 1
=
13.4458 = 1.017 14 - 1
x1 = 7.881
s1=1.017
x 2 = 8.430
=STDEV(A2:A15)
Step 4: Conclusion--Because the calculated t = -1.437 is greater than the left critical t = -2.160 (and in the nonrejection region), do not reject H0 at the 0.05 significance level. There is not a significant difference in the mean red blood cell mass of the flight rats and the control rats at the 0.05 level.
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The red blood cell mass problem for flight and control rats was calculated in Excel using a "tTest: Two-Sample Assuming Unequal Variances."--the Excel output is shown at the bottom of this page. Notice that df=26 in the Excel output versus df=13 in the hand calculations above. Your textbook (p. 525) explains that a more accurate way to calculate df is using the following df formula, which is th...