Question

a credit score is used by credit agencies (such as mortgage companies and banks) to assess the creditworthiness of individuals. values range from 300 to 850, with a credit score over 700 considered to be a quality credit risk. according to a survey, the mean credit score is 708.8. a credit analyst wondered whether high - income individuals (incomes in excess of $100,000 per year) had higher credit scores. he obtained a random sample of 45 high - income individuals and found the sample mean credit score to be 727.5 with a standard deviation of 83.9. conduct the appropriate test to determine if high - income individuals have higher credit scores at the \\( \alpha=0.05 \\) level of significance

state the null and alternative hypotheses.

\\( h_{0}: \mu=708.8 \\)
\\( h_{1}: \mu>708.8 \\)
(type integers or decimals. do not round.)

identify the t - statistic.

\\( t_{0}=\square \\) (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for t - statistic

The formula for the t - statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Substitute the given values into the formula

We are given \(\bar{x} = 727.5\), \(\mu=708.8\), \(s = 83.9\), and \(n = 45\).
First, calculate the denominator \(s/\sqrt{n}=\frac{83.9}{\sqrt{45}}\approx\frac{83.9}{6.7082}\approx12.507\).
Then, calculate the numerator \(\bar{x}-\mu=727.5 - 708.8=18.7\).
Now, find \(t=\frac{18.7}{12.507}\approx1.50\).

Answer:

\(t_0 = 1.50\)