Simple random samples of high-interest mortgages and low-interest mortgages were
ID: 3362386 • Letter: S
Question
Simple random samples of high-interest mortgages and low-interest mortgages were obtained. For the 15 high-interest mortgages, the borrowers had a mean FICO score of 472 and a standard deviation of 25. For the 22 low-interest mortgages, he borrowers had a mean FICO credit score of 485 and a standard deviaiton of 30. Test the claim that the mean FICO score of borrowers with high-interest mortgages is lower than the mean FICO score of borrowers with low-interest mortgages at the .10 significance level.
Claim: Select an answer 1 2 1 > 2 1 2 1 2 p 1 p 2 p 1 > p 2 1 < 2 p 1 < p 2 p 1p 2 p 1 = p 2 p 1 p 2 1 = 2 which corresponds to Select an answer H1: p 1 > p 2 H1: p 1 < p 2 H1: 1 > 2 H0: 1 2 H1: p 1p 2 H0: 1 2 H0: 1 = 2 H1: 1 2 H1: 1 < 2 H0: p 1 p 2
Opposite: Select an answer 1 < 2 1 2 p 1 p 2 p 1 p 2 p 1 = p 2 p 1 > p 2 1 > 2 1 2 1 2 p 1p 2 1 = 2 p 1 < p 2 which corresponds to Select an answer H1: 1 > 2 H0: 1 2 H1: p 1 p 2 H1: p 1 <= p 2 H1: 1 < 2 H0: 1 2 H0: p 1p 2 H1: 1 2 H0: 1 = 2 H0: p 1 > p 2 H1: p 1 = p 2
The test is: Select an answer right-tailed left-tailed two-tailed
The test statistic is: tt = Select an answer -0.934 -1.55 -1.054 -1.431 -1.173
The critical value is: Select an answer -1.701 -1.345 -1.635 -1.2 -1.59
Based on this we: Select an answer Cannot determine anything Fail to reject the null hypothesis Accept the null hypothesis Reject the null hypothesis
Conclusion There Select an answer does does not appear to be enough evidence to support the claim that the mean FICO score of borrowers with high-interest mortgages is lower than the mean FICO score of borrowers with low-interest mortgages.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: High> Low
Alternative hypothesis: High < Low
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 9.09
DF = 35
The critical value of t = -1.345
t = [ (x1 - x2) - d ] / SE
t = - 1.43
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.
The observed difference in sample means (10) produced a t statistic of - 1.43. We use the t Distribution Calculator to find P(t < - 1.43)
Therefore, the P-value in this analysis is 0.081
Interpret results. Since the P-value (0.081) is less than the significance level (0.10), we have to reject the null hypothesis.
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