. Question 4 On a peer-to-peer (P2P) lending website, borrowers complete an approval scoring form that lenders use to assess creditworthiness. Lenders generally believe that borrowers who score at least 71 do not default on loans. Consequently, borrowers are rated A if their overall score is at least 71, otherwise they are rated B. A reasonably large sample of real borrower data was collected: i. among those that did not default on their loans, initial approval scores were normally distributed with a mean of 75 and a standard deviation of 8.1. ii. among those that defaulted on their loans, initial approval scores were normally distributed with a mean of 64.3 and a standard deviation of 11. Report each answer as a decimal (not percent) accurate to at least 4 decimal places. Answers from software or from rounded z-scores (to 2 decimal places) are accepted. 1. What proportion of borrowers that: a) defaulted were initially rated A? bj did not default were initially rated B? c) defaulted were miscategorized initially? d) did not default were miscategorized initially? 2. Among those that defaulted, what is the probability that a borrower: a) scored above 77.1? bj scored below 70.4 or above 80.5? c) was rated B and scored above 69.37 d) was rated A and scored below 77.1? e) was rated A given scored above 64.3? f) scored below 64.3 given rated B?. Question 7 Consider a population with a mean / = 227 and standard deviation o = 63. Suppose random samples of size n = 83 are selected from this population. 1. a) What is the mean of the distribution of the sample mean? HI = b) What is the standard error of the mean? (2 decimal places) In the following questions, round the standard error and z-scores to exactly 2 (two) decimal places before determining probabilities. Report probabilities accurate to at least 4 decimal places. 2. What is the probability that a randomly selected sample mean is: a) greater than 234.1? b) less than 238.87 c) greater than 234.1 and less than 238.8? d) greater than 234.1 given less than 238.8? e) greater than 234.1 or greater than 238.87 3. Between what values would you expect to find the middle 96% of the sample means? Round to the nearest integer. Place the smaller value in the first box and the larger value in the second box. Between and 4. Why are we able to use the normal distribution in the calculations above? O Because the sample mean is large enough O Becasuse the standard error is large enough O Because the sample size is large enough O Because the original population is normal. Question & When constructing the confidence interval a population mean, assuming the population standard deviation is known, we use the following formula: Critical Value Standard Error Sample Meen x+ Za/2 Margin of Error Vn The table below contains information about two confidence intervals. Use the given information to determine the missing values: Round answers accurate to 2 decimal places. Interval 1 Interval 2 Confidence 87.40% Level Critical Value 2.75 (za/2) Sample 121.50 111.60 Mean Standard 6.70 Error Margin of Error Lower Limit 100.43 Upper Limit copyable table Interval 1 Interval 2 Confidence Level 87.40% Critical Value (z) 2.75 Sample Mear 121.50 111.60 Standard 6.70 Margin of Error Lower Limit 100.43 Upper Limit. Question 13 Using two different confidence levels, a researcher constructs two confidence intervals for a population mean based on a sample of size n = 21. The standard error of the mean is 5.2372. Assume the population is normally distributed and the population standard deviation is unknown. The confidence intervals are: Interval 1: 80.36 45 Ou 41.7 b. At a = 0.095, determine the critical value (2 decimal places): Critical Value = c. The value of the test statistic (2 decimal places) = d. At a = 0.095, select the decision/conclusion: Reject the null hypothesis. There is sufficient evidence to conclude that the new efficiencies introduced make the completion of the process faster._- Assume the hourly wage rate for a fast-growing online job is normally distributed with a mean of p = $21.00. Suppose a random sample of n. = 25 workers revealed a mean wage of E: = $20.00 and a standard deviation of a = $1.10. At or = 0.025. is there enough Evidence to conclude that the mean hourly wage is less than $21.00? a. Select the appropriate alternative hypothesis, H1 or H\": Op. % $21.00 Op. '3: $21.00 Up. 9E $20.00 op. o: $20.00 0p. :1: $20.00 0p. :1: $21.00 b. At or = 0.025, select the critical valuels}: c. The value of the test statistic {3 decimal places] = D d. At or = 0.025, select the decisionl'oonolusion: enough evidence to conclude that the mean hourly\":r wage is less than $21.00. . Question 20 In testing the hypothesis Ho : / 2 40 at a = 0.1, a researcher calculated the test statistic as z = - 0.84 and made the appropriate decision based on this statistic. The true population mean was later found to be 32. Select all true statements about the researcher's decision below. The researcher: O made a correct decision Ofailed to reject the null hypothesis O committed both Type I and Type II errors Oaccepted the null hypothesis Orejected the null hypothesis O committed a Type | error O committed a Type ll error O made a mistake in the calculations