Question 1:
A random sample of the birth weights of 186 babies has a mean of 3103g and a standard deviation of 696g (based on data from ''Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure,'' by Singer et al., Journal of the American Medical Association, Vol. 291, No. 20. These babies were born to mothers who did not use cocaine during their pregnancies. Further, a random sample of the birth weights of 190 babies born to mothers who used cocaine during their pregnancies has a mean of 2700g and a standard deviation of 645g.
a) Set up the null and alternative hypotheses to test the claim that both samples are from the populations having the same standard deviation.
Null Hypothesis (Ho): σ1 = σ2
Alternative Hypothesis (Ha): σ1 ≠ σ2
Where σ1 = population standard deviation of mothers who did not use cocaine during their pregnancies and σ2 = population standard deviation of mothers who used cocaine during their pregnancies
b) Perform the hypothesis test you set up in a) with the significance level a = 0.05.
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F Test for Differences in Two Variances
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Data
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Level of Significance
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0.05
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Larger-Variance Sample
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Sample Size
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186
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Sample Variance
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484416
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Smaller-Variance Sample
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Sample Size
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190
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Sample Variance
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416025
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Intermediate Calculations
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F Test Statistic
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1.1644
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Population 1 Sample Degrees of Freedom
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185
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Population 2 Sample Degrees of Freedom
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189
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Two-Tail Test
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|
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Upper Critical Value
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1.3329
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p-Value
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0.2986
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Do not reject the null hypothesis
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c) Using the results of b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.
Since the p-value of the test is bigger than 0.05 level of significance so we will not be able to reject the null hypothesis and conclude that cocaine use does not appear to affect the birth weight of a baby.
Question 2:
Simple random samples of high-interest (5.36%) mortgages and low-interest (3.77%) mortgage were obtained. For the 40 high-interest mortgages, the borrowers had a mean FICO credit score of 594.8 and standard deviation of 12.2. For the 40 low-interest mortgages, the borrowers had a mean FICO credit score of 785.2 and standard deviation of 16.3.
a) Use a 0.01 significance level to test the claim that the mean FICO score of borrowers with high-interest mortgage is lower than the mean FICO score of borrowers with low-interest mortgage.
Null Hypothesis (Ho): µ1 = µ2
Alternative Hypothesis (Ha): µ 1 < µ2
Where µ1 = population mean FICO score of borrowers with high-interest mortgage and µ2 = population mean FICO score of borrowers with low-interest mortgage
b) Does the FICO credit rating score appear to affect mortgage payments? If so, how?
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Pooled-Variance t Test for the Difference Between Two Means
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(assumes equal population variances)
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Data
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Hypothesized Difference
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0
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Level of Significance
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0.01
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Population 1 Sample
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Sample Size
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40
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Sample Mean
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594.8
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Sample Standard Deviation
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12.2
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Population 2 Sample
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Sample Size
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40
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Sample Mean
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785.2
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Sample Standard Deviation
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16.3
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Intermediate Calculations
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Population 1 Sample Degrees of Freedom
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39
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Population 2 Sample Degrees of Freedom
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39
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Total Degrees of Freedom
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78
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Pooled Variance
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207.2650
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Standard Error
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3.2192
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Difference in Sample Means
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-190.4000
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t Test Statistic
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-59.1451
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Lower-Tail Test
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Lower Critical Value
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-2.3751
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p-Value
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0.0000
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Reject the null hypothesis
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Question 3:
A random sample of the birth weights of 186 babies has a mean of 3103g and a standard deviation of 696g (based on data from "Cognitive Outcomes of Preschool Children with Prenatal Cocaine Exposure," by Singer et al., Journal of the American Medical Association, Vol. 291, No. 20). These babies were born to mothers who did not use cocaine during their pregnancies. Further, a random sample of the birth weights of 190 babies born to mothers who used cocaine during their pregnancies has a mean of 2700g and a standard deviation of 645g.
a) The birth weights of babies are known normally distributed. Use a 0.05 significance level to test the claim that both samples are from populations having the same standard deviation.
Null Hypothesis (Ho): σ1 = σ2
Alternative Hypothesis (Ha): σ1 ≠ σ2
Where σ1 = population standard deviation of mothers who did not use cocaine during their pregnancies and σ2 = population standard deviation of mothers who used cocaine during their pregnancies
a) Using your finding in (a), construct a 90% confidence interval estimate of the difference between the mean birth weight of a baby born to mothers who did not use cocaine and that of a baby born to mothers who used cocaine during their pregnancies.
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F Test for Differences in Two Variances
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Data
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Level of Significance
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0.05
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Larger-Variance Sample
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Sample Size
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186
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Sample Variance
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484416
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Smaller-Variance Sample
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|
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Sample Size
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190
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Sample Variance
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416025
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|
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Intermediate Calculations
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F Test Statistic
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1.1644
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Population 1 Sample Degrees of Freedom
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185
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Population 2 Sample Degrees of Freedom
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189
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Two-Tail Test
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|
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Upper Critical Value
|
1.3329
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p-Value
|
0.2986
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Do not reject the null hypothesis
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b) Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.
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Confidence Interval Estimate
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for the Difference Between Two Means
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Data
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Confidence Level
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90%
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Intermediate Calculations
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Degrees of Freedom
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374
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t Value
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1.6489
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Interval Half Width
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114.0778
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Confidence Interval
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Interval Lower Limit
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288.9222
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Interval Upper Limit
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517.0778
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c) Using your finding in (b), does cocaine use appear to affect the birth weight of a baby? Substantiate you conclusion.
We can see that 0 does not lie inside the 90% confidence interval so we can conclude that cocaine use appear to affect the birth weight of a baby. We can see that the mean weight of the baby is high whose mother does not use cocaine as compare to mothers who used cocaine during their pregnancies.
Question 4:
A large discount chain compares the performance if its credit managers in Ohio and Illinois by comparing the mean dollars amounts owed by customers with delinquent charge accounts in these two states. Here a small mean dollar amount owed is desirable because it indicates that bad credit risks are not being extended large amounts of credit. Two independent, random samples of delinquent accounts are selected from the populations of delinquent accounts in Ohio and Illinois, respectively. The first sample, which consists of 15 randomly selected delinquent accounts in Ohio, givens a mean dollar amount of $524 with a standard deviation of $38. The second sample, which consists of 20 randomly selected delinquent accounts in Illinois, gives a mean dollar amount of $473 with a standard deviation of $22.
a) Assuming that the normality assumption, test to determine if the population variances are equal with a =0.05.
Null Hypothesis (Ho): σ1 = σ2
Alternative Hypothesis (Ha): σ1 ≠ σ2
Where σ1 = population standard deviation of Ohio and σ2 = population standard deviation of Illinois
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F Test for Differences in Two Variances
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|
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Data
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Level of Significance
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0.05
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Larger-Variance Sample
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|
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Sample Size
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15
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Sample Variance
|
1444
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Smaller-Variance Sample
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Sample Size
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20
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Sample Variance
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484
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Intermediate Calculations
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F Test Statistic
|
2.9835
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Population 1 Sample Degrees of Freedom
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14
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Population 2 Sample Degrees of Freedom
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19
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Two-Tail Test
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Upper Critical Value
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2.6469
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p-Value
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0.0283
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Reject the null hypothesis
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b) Test with a =0.05 whether there is a difference between the population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and Illinois.
Null Hypothesis (Ho): µ1 = µ2
Alternative Hypothesis (Ha): µ 1 ≠ µ2
Where µ1 = population mean dollar amounts owed by consumers with delinquent charge accounts in Ohio and µ2 = population mean dollar amounts owed by consumers with delinquent charge accounts in Illinois
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Separate-Variances t Test for the Difference Between Two Means
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(assumes unequal population variances)
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Data
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Hypothesized Difference
|
0
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Level of Significance
|
0.05
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Population 1 Sample
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|
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Sample Size
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15
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Sample Mean
|
524
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Sample Standard Deviation
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38.0000
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Population 2 Sample
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Sample Size
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20
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Sample Mean
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473
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Sample Standard Deviation
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22.0000
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Intermediate Calculations
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Numerator of Degrees of Freedom
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14512.2178
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Denominator of Degrees of Freedom
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692.7711
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Total Degrees of Freedom
|
20.9481
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Degrees of Freedom
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20
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Standard Error
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10.9757
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Difference in Sample Means
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51.0000
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Separate-Variance t Test Statistic
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4.6466
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Two-Tail Test
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|
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Lower Critical Value
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-2.0860
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Upper Critical Value
|
2.0860
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p-Value
|
0.0002
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Reject the null hypothesis
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c) Assuming that the normality assumption and the condition you checked in a) hold, calculate a 95 percent confidence interval for the difference between the mean dollar amounts owed in Ohio and Illinois. Based on this interval, do you think that these mean dollar amounts differ in a practically important way?
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Confidence Interval Estimate
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for the Difference Between Two Means
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|
|
|
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Data
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|
Confidence Level
|
95%
|
|
|
|
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Intermediate Calculations
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|
Degrees of Freedom
|
33
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t Value
|
2.0345
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Interval Half Width
|
20.7463
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|
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Confidence Interval
|
|
Interval Lower Limit
|
30.2537
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Interval Upper Limit
|
71.7463
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