Can you give detailed explanation as to where all the numbers come from I have all the answers but not sure where all the numbers are coming from in the formula. Thanks again! These are practice questions that already have answers just need to better understand them

A highway patrol officer believes that the average speed of cars traveling over a certain stretch of highway exceeds the posted limit of 70 mph. The speeds of a random sample of 100 cars were recorded. The sample average and the sample standard deviation were found as 71 mph and 5 mph, respectively. The officer develops the following test: Ho: US 70 Ha: H > 70. Answer Questions 2 through 4 based on this information. Question 2 The test statistic is: A) 1,82 B) 2.0 C) 2.5 D) 0.05 E) None of the above OPRE 6301 Exam 3 Practice Questions SOLUTION: (B) In this case, population standard deviation is not known. Therefore, to = X-Ho - 71- s/Vn 5/V100 = = 2 Question 3 When level of significance is 0.05 (a=0.05), what is the critical value that defines the rejection region? A) 1.66 B) 2.626 C) 1.645 D) 0.6776 E) 1.96 SOLUTION: (A) The rejection region is to > tan-1. That is, the critical value is to.05,99=1.66Question 4 At .05 level of significance, it can be concluded that the average speed of cars A) not significantly greater than 100 B) significantly greater than 100 C) not significantly less than 100 D) significantly less than 100 E) None of the above SOLUTION: (B) The rejection region is to > tan-1. The critical value is to.05,99=1.66. Since to = 2> tan-1 = 1.66, we reject the null hypothesis, and conclude that the average speed of cars is significantly greater than 100. 2 OPRE 6301 Exam 3 Practice Questions Consider the following hypothesis test when a = 0.05. Ho: H1 - 12 = 0 Ha:My - H2 0 The following results are from independent samples taken from two populations. Sample 1 Sample 2 n = 35 n2 = 40 X1 = 13.6 X2 = 10.1 $1 = 5.2 $2 = 8.5 Assuming that population variances are equal, answer Questions 5 through 9 based on the above two samples.Question 5 The pooled estimate of the variance (s?), is: A) 7.16 B) 6.85 C) 46.93 D) 2.928 E) 51.19 SOLUTION: (E) (n1 - 1)s + (n2 -1)s, 34 x 5.22 + 39 x 8.52 - = 51. 19 n1 + n2 - 2 73 Question 6 The test statistic is: A) 0.2954 B) 2.114 C) 0.489 D) 0.2147 E) 7.61 OPRE 6301 Exam 3 Practice Questions SOLUTION: (B) to = (21-12)-Ho_(13.6-10.1)-0 _2.114 1+1 51.19 141 Spam1 12Question 7 The rejection region is: A) -1.666 s to $ 1.666 B) to >1.666 C) to 1.666 E) Zo 1.666 SOLUTION: (D) The rejection region is Itol > tan+na-2 to.05,35-40-2= to.05,73=1.666. Therefore, our rejection region is to 1.666. Question 8 What is the p-value? A) 0.981 B) 0.038 C) 0.019 D) 0.95 E) 0.05 SOLUTION: (B) Because to O, p-value=2xP(to.05,73>2.114)=0.038Question 9 What is your conclusion about the hypothesis test in consideration? A) , is greater than #2 B) , is no less than uz C) u, is not equal to #2 D) Current evidence is not enough to reject null E) u, is less than #2 SOLUTION: (C) Since p-value=0.038 3.747Question 15 What does the analyst conclude as a result of the hypothesis test mentioned in Question (13)? A. There is enough evidence to infer a linear relationship B. There is not enough evidence to infer a linear relationship C. There may be a linear relationship but the current sample does not provide enough evidence D. The hypothesis test is inconclusive E. None of the above SOLUTION: (A) Question 16 At a 98% confidence level, what would be the analyst's estimate for y in terms of a prediction interval given that x=7? A. 283.469+31.408 B. 283.4691275.42 C. 283.469+220.403 D. 283.469+247.342 E. None of the above SOLUTION: (B) 17.5_17.5-35 n-1 6-1 5 The following gives us a prediction interval for x,=7. y tta _ 2n-252 1+ 1 (x - 1)? n (n - 1)s2 Note that, we have found s:= 53.81, previously. Furthermore, when 1- a=0.98, we have a=0.02. This leads to ta _, = to.01,4=3.747.A B C D E F 1 SUMMARY OUTPUT 2 3 Rogrossion Statistics 4 Multiple R 0.5926 5 R Square 0.3511 6 Adjusted R Square 0.3352 7 Standard Error 6.99 8 Observations 126 10 ANOVA 11 SE MS Significance F 12 Regression 3 3227.6 1075.9 22.01 1 86E-11 13 Residual 122 5964.5 18.89 14 Total 125 2192.1 15 16 Coefficients |Standard Error Star p-value 17 Intercept -1.97 9 55 -0 206 0.8309 18 Minor HR 0.606 0.0871 7 64 5 46F-12 19 Age 0.130 0.524 0.259 0.7901 20 Years Pro 1.18 0.671 1.75 0.0819 Answer Questions 17 through 21 for the general manager's regression model. Question 17 What is the estimated regression equation? A) y = -1.97 + 0.666x, + 0.136*2 + 1.18x3 B) y = -1.97x, + 0.666x2 + 0.136x3 + 1.18 C) y = -1.97xo + 0.666x1 + 0.136x2 + 1.18x3 D) y = -1.97 + 0.666x, + 0.136x2 E) = -1.97 + 0.666x1 SOLUTION: (A) y = -1.97 + 0.666x, + 0.136x2 + 1.18x3Question 18 Which one of the following is a correct statement? A) The unexplained variation in y is 3227.6 B) The explained variation in y is 2736.9 C) The portion of the variability in y that can be explained by the suggested regression is 0.3511 D) There are 125 observations in the data set E) None of the above SOLUTION: The portion of the variability in y that can be explained by the suggested regression is 0.3511 (R-squared) Question 19 The general manager thinks that R-Square in this model is less than expected. He further performs the following hypothesis test for each independent variable: Ho: B, = 0 H,: B, + 0. He uses a=0.05. As a result of this test, which of the independent variables can s/he eliminate? A) X1 B) x2 and x3 C) x1 and X2 D) X1 and x3 E) None SOLUTION: (B)