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 703.1. 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 35 high-income individuals and found the sample mean credit score to be 716.6 with a standard deviation of 80.1. Conduct the appropriate test to determine if high-income individuals have higher credit scores at the alphaequals0.05 level of significance.
Accepted Solution
A:
Answer:[tex]t=\frac{715.4-703.8}{\frac{82.7}{\sqrt{40}}}=0.887[/tex] [tex]p_v =P(t_{(39)}>0.887)=0.1902[/tex]If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis. And we can conclude that the true mean i not significantly different from 703.8 Step-by-step explanation:Data given and notation [tex]\bar X=715.4[/tex] represent the sample mean [tex]s=82.7[/tex] represent the sample standard deviation for the sample [tex]n=40[/tex] sample size [tex]\mu_o =703.8[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test. t would represent the statistic (variable of interest) [tex]p_v[/tex] represent the p value for the test (variable of interest) State the null and alternative hypotheses. We need to conduct a hypothesis in order to check if the mean is higher than 703.8, the system of hypothesis would be: Null hypothesis:[tex]\mu \leq 703.8[/tex] Alternative hypothesis:[tex]\mu > 703.8[/tex] If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by: [tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1) t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value". Calculate the statistic
We can replace in formula (1) the info given like this: [tex]t=\frac{715.4-703.8}{\frac{82.7}{\sqrt{40}}}=0.887[/tex] P-value
The first step is calculate the degrees of freedom, on this case: [tex]df=n-1=40-1=39[/tex] Since is a right tailed test the p value would be: [tex]p_v =P(t_{(39)}>0.887)=0.1902[/tex]Conclusion If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to fail reject the null hypothesis. And we can conclude that the true mean i not significantly different from 703.8