Bankrate.com reports mortgage loan interest rates for 30-year and 15-year fixed-

Question

Bankrate.com reports mortgage loan interest rates for 30-year and 15-year fixed-rate mortgage loans for 9 randomly selected Willamette Valley lending institutions. Assuming normality, a financial analyst would like to determine whether there is a difference between mean 30-year rates and mean 15-year rates. a) State null hypothesis and alternative hypothesis. Ho: Ha: b) Calculate the value of the test statistics. Specify the test statistic ("z" or t") clearly. c) Find the p-value using the appropriate table. Draw the curve and highlight the corresponding area. d) At the .05 significance level, do you reject or retain the null hypothesis? Circle the correct answer and explain using the p-value found in part c). Reject Ho/Retain Ho Why? c) Interpret the result, this includes the weight of evidence, the conclusion, and the significance level.

Explanation / Answer

Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: 1 - 2 = 0 , that s the two means are equal.
Alternative hypothesis: 1 - 2 0, that is the two means are different.

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = sqrt[(0.1452/9) + (0.183962/9)] = 0.0781
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (0.1452/9 + 0.183962/9)2 / { [ (0.1452 / 9)2 / (8) ] + [ (0.183962 / 9)2 / (8) ] }
DF = 0.0000371643 / 0.0000024495 = 15.17

t = [ (x1 - x2) - d ] / SE = [ (6.7189 - 6.4841) - 0 ] / 0.0781 = 3.01

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 the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability

We use the t Distribution Calculator to find P(t < 3.01)

The P-Value is 0.008704.
The result is significant at p < 0.05.

Interpret results. Since the P-value (0.008704) is less than the significance level (0.05), we cannot accept the null hypothesis

Conclusion. Reject the null hypothesis. We have insufficient evidence to prove the claim, that the difference between the mean 30 years rate and 15 years rate are equal.

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