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# Lenders tighten or loosen their standards for issuing credit as economic conditions change. One of the...

 Lenders tighten or loosen their standards for issuing credit as economic conditions change. One of the criteria lenders use to evaluate the creditworthiness of a potential borrower is her credit risk score, usually a FICO score. FICO scores range from 300 to 850. A consumer with a high FICO score is perceived to be a low credit risk to the lender and is more likely to be extended credit than a consumer with a low score. A credit card represents a line of credit, because the credit card holder obtains a loan whenever the card is used to pay for a purchase. A study of credit card accounts opened in 2002 found a mean FICO score for the credit card holder (at the time the card was issued) of 731 and a standard deviation of 76. [Source: Sumit Agarwal, John C. Driscoll, Xavier Gabaix, and David Laibson, “Learning in the Credit Card Market,” Working Paper 13822, National Bureau of Economic Research (NBER), February 2008.] You conduct a hypothesis test to determine whether banks have loosened their standards for issuing credit cards since 2002. You collect a random sample of 100 credit cards issued during the past 6 months. The sample mean FICO score of the credit card holders (at the time their cards were issued) is x̄x̄ = 715. Assume that the standard deviation of the population of FICO scores for credit cards issued during the past 6 months is known to be σ = 76, the standard deviation from the NBER study. Let µ equal the true population mean FICO score for consumers issued credit cards in the past 6 months. You should formulate the null and alternative hypotheses as: H₀: µ = 731, H11 : µ < 731 H₀: µ = 731, H11 : µ > 731 H₀: x̄x̄ = 731, H11 : x̄x̄ < 731 H₀: µ < 731, H11 : µ = 731 If the null hypothesis is true as an equality, the sampling distribution of x̄x̄ is approximated by selector 1    a t a standard normal a normal a binomial an unknown distribution with selector 2    a mean of 0 an unknown mean a mean of 715 a mean of 731 and a standard deviation of selector 3    76 1.0 unknown 7.6 . The value of the standardized test statistic is selector 1    z = 16 t = –2.11 z = –2.11 z = 715 . Use the Distributions tool to help you answer the questions that follow. Normal Distribution Mean = 730 Standard Deviation = 8.5 You conduct the hypothesis test using a significance level of α = 0.05. Use the tool to develop the rejection region for your test. According to the critical value approach, when do you reject the null hypothesis? Reject H1 if z ≤ –1.645 Reject H₀ if z ≥ –2.11 Reject H₀ if z ≤ –1.645 Reject H₀ if z ≤ –1.96 or z ≥ 1.96 The p-value is selector 1    16 0.9826 –2.11 0.0174 . Using the critical value approach, the null hypothesis is selector 1    not rejected rejected , because selector 2    –2.11 < –1.645 7.6 > –1.645 0.0174 < 0.05 715 < 731 . Using the p-value approach, the null hypothesis is selector 3    not rejected rejected , because selector 4    0.0174 < 0.05 0.9826 > 0.05 0.0143 < 0.05 –2.11 < –1.645 . Therefore, you selector 5    cannot can conclude that banks have loosened their standards for issuing credit cards since 2002.

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