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. A magazine claims that the mean amount spent by a customer at burger Stop is greater than the mean amount spent by a customer

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. A magazine claims that the mean amount spent by a customer at burger Stop is greater than the mean amount spent by a customer at Fry World. The results do samples of customer transactions for the two fast food restaurants are shown in the table. At 5% can you support the magazine's claim. Assume that population are normally distributed, and population variances are equal. Burger Stop Fry World = $5.46 *2 = $5.12 $1 = $0.89 $2 = $0..79 n1 = 22 n2 = 30 a) Identify the claim and state null and alternative hypothesis. b) Find the standardized test statistic and corresponding critical value. c) At 5% significance level, decide whether to reject or fail to reject the null hypothesis. d) Interpret the decision in the context of the original claim. 2. The table shows the credit scores for 12 randomly selected adults who are considered high risk barrowers before and two years after they attend a personal finance seminar. Adult 1 2 3 4 5 6 7 8 9 10 11 1 12 Credit score 608 620 610 650 640 680 655 602 644 656 632 664 (before seminar) Credit score (after 646 692 715 669 725 786 700 650 660 650 680 702 seminar) Based on the information provided in the output below answer following questions: t-Test: Paired Two Sample for Means Variable I Variable 2 Mean 638.417 689.583 Variance 597.720 1626.265 Observations 12 12 Pearson Correlation 0.509 Hypothesized Mean Difference df 11 t Stat -5.073 P(T It Intercept 0.64031 0.22922 2.793 0.00728 log(bigmac2003) 0.80293 0.06709 11.967 F Regression 2322 2322 8.71 0.032 Residual 1334 267 Total 3656 A) This cannot be determined from the information given. B) 0.3649 C) 0.5745 D) 0.6351 The regression output for this simple linear regression model is given below: Regression Statistics: R = 0.6669 Adj. R= = 0.6613 S = 12 5.. = 20.6 S. = 11.3 y = 96.2 Y = 57.1 ANOVA Source DF Sum of Mean Square F Value Prob > F Squares Regression 1 17003.4 17003.4 118.13 t Intercept 181.1443 7.9652 22.74

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