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A loan officer compares the interest rates for 4848-month fixed-rate auto loans and 4848-month variable-rate auto loans. Two independent, random samples of auto loan rates are selected. A sample of eight 4848-month fixed-rate auto loans had the following loan rates:

4.29%3.75%3.50%3.99%3.75%3.99%5.40%4.00%\begin{array}{llll} 4.29 \% & 3.75 \% & 3.50 \% & 3.99 \% \\ 3.75 \% & 3.99 \% & 5.40 \% & 4.00 \% \end{array}

while a sample of five 4848-month variable-rate auto loans had loan rates as follows:

3.59%2.75%2.99%3.00%\begin{array}{llll} 3.59 \% & 2.75 \% & 2.99 \% & 3.00 \% \\ \end{array}

Figure 10.710.7 gives the Excel output of using the equal variances procedure to test the hypotheses you set up in part aa. Assuming that the normality and equal variances assumptions hold, use the Excel output and critical values to test these hypotheses by setting α\alpha equal to .10.10, .05.05, .01.01, and .001.001. How much evidence is there that the mean rates for 4848 month fixed and variable rate auto loans differ?

Figure 10.710.7 Excel Output of Testing the Equality of Mean Loan Rates for Fixed and Variable 4848-Month Auto Loans t-Test: Two-Sample Assuming Equal Variances

 Fixed-Rate (%)  Variable-Rate (%)  Mean 4.08382.966 Variance 0.33760.1637 Observations 85 Pooled Variance 0.2744 Hypothesized Mean Difference 0 df 11 t Stat 3.7431 P(T <=t) one-tail 0.0016 t Critical one-tail 1.7959 P(T <=t) two-tail 0.0032 t Critical two-tail 2.2010\begin{array}{lrr} & \text{ Fixed-Rate (\\\%) } & \text{ Variable-Rate (\\\%) } \\ \text { Mean } & 4.0838 & 2.966 \\ \text { Variance } & 0.3376 & 0.1637 \\ \text { Observations } & 8 & 5 \\ \text { Pooled Variance } & 0.2744 & \\ \text { Hypothesized Mean Difference } & 0 & \\ \text { df } & 11 \\ \text { t Stat } & 3.7431 \\ \text { P(T }<=\mathrm{t}) \text { one-tail } & 0.0016 \\ \text { t Critical one-tail } & 1.7959 \\ \text { P(T }<=\mathrm{t}) \text { two-tail } & 0.0032 & \\ \text { t Critical two-tail } & 2.2010 \end{array}

Question

A loan officer compares the interest rates for 4848-month fixed-rate auto loans and 4848-month variable-rate auto loans. Two independent, random samples of auto loan rates are selected. A sample of eight 4848-month fixed-rate auto loans had the following loan rates:

4.29%3.75%3.50%3.99%3.75%3.99%5.40%4.00%\begin{array}{llll} 4.29 \% & 3.75 \% & 3.50 \% & 3.99 \% \\ 3.75 \% & 3.99 \% & 5.40 \% & 4.00 \% \end{array}

while a sample of five 4848-month variable-rate auto loans had loan rates as follows:

3.59%2.75%2.99%3.00%\begin{array}{llll} 3.59 \% & 2.75 \% & 2.99 \% & 3.00 \% \\ \end{array}

Calculate a 9595 percent confidence interval for the difference between the mean rates for fixed- and variable-rate 4848-month auto loans. Can we be 9595 percent confident that the difference between these means exceeds .4.4 percent? Explain.

Solution

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When we want to compute a 100(1α)%100(1-\alpha)\% confidence interval for μ1μ2\mu_1-\mu_2 , we have to use this formula:

[(xˉ1xˉ2)±tα/2sp2(1n1+1n2)].\left[ (\=x_1-\=x_2) \pm t_{\alpha/2} \sqrt{s_p^2 \left(\frac{1}{n_1}+\frac{1}{n_2}\right)} \right].

We use this formula when we know that variances are equal.

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