Need help with screenshots 1 through 7. All information needed can be found below. Note: Each screenshot includes an example (reference), the reason for this is because each screenshot includes multiple parts. Please complete every single part if possible. If you need any more information, do not hesitate to let me know. Thank you!

1, sample Automobile collision insurance is used to pay for any claims made against the driver in the event of an accident. This type of insurance will typically pay to repair any assets that your vehicle damages. Arandom sample of40 collision claims of 20- to 24year-old drivers results in a mean claim of $4590 with a standard deviation of 32315. An independent random sample of 40 collision claims of 30- to 59-year-old drivers results in a mean claim M53669 with a standard deviation of $2029. Using the concept of hypothesis testing, determine if a higher insurance premium should be paid by 20- to 24-year-old drivers. Use a u = 0.01 level of signicance, and let population 1 be 20- to 24~year old drivers and population 2 be 30- to 59-year old drivers. Complete parts (a) through (e) below. (a) Collision claims tend to be skewed right. Why do you think this is the case? *A. There are a few very large collision claims relative to the majority of claims. 'x B. There are no very large collision claims. C. There are many large collision claims relative to the majority of claims. (b) What type of test should be used? A. A hypothesis test regarding the difference between two population proportions from independent samples * B. A hypothesis test regarding the dilierence of two means using Welch's approximate t J'C. Ahypothesis test regarding two population standard deviations D. A hypothesis test regarding the difference of two means using a matched-pairs design (c) Determine the null and alternative hypotheses. V V Ho: P1 = P2 V V H1: P1 > P2 (d) Use technology to calculate the P-value. 0.031' (Round to three decimal places as needed.) (9) Draw a conclusion based on the hypothesis test. Choose the correct answers below. There is not sufcient evidence to reject the null hypothesis that the mean claims of 20- to 24-yearold drivers and 30- to 59-yearold drivers are equal because Pvalue >' a. Help Me Solve This View an Example Get More Help A Aulomobile collision insurance is used to pay (or any claims made against the driver in the event of an accident, This type of insurance will typically pay to repair any assets that your vehicle damages. A random sample of 40 collision claims of 20- to 24year-old drivers results in a mean claim of $4580 with a standard deviation 0! $2291. An independent random sample of 40 collision claims of 30- to 59-year-old drivers results in a mean claim of $3650 with a standard deviation of $2014. Using the concept of hypothesis testing, determine il a higher insurance premium should be paid by 20- to 24-year~old drivers. Use a u = 0.01 level of signicance, and let population 1 be 20- to 24-year old drivers and population 2 be 30- to 59-year old drivers. Complete pans (a) through (e) below. (a) Collision claims tend to be skewed right. Why do you lhink this is the case? 0 A. There are a few very large collision claims relative to the majority of claims. 0 B. There are no very large collision claims. 0 C. There are many large collision claims relative to the majority of claims. Help Me Solve This View an Example Get More Help A Clear All 2, sample Automobile collision insurance is used to pay for any claims made against the driver in the event of an accident. This type of insurance will typically pay to repair any assets that your vehicle damages. Complete parts (a) and (b). Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Click here to view the table of critical F-values (page 1), Click here to view the table of critical F-values (page 2). Click here to view the table of critical F-values (page 3). Click here to view the table of critical F-values (page 4), (a) Collision claims tend to be skewed right. Why do you think this is the case? O A. Most claims are moderate with similar numbers of low value and high value claims. B. Most claims are either low value or high value, with not many claims for moderate values. O C. Most claims are high, but there are some claims that are much lower than the average. D. Most claims are moderate, but there are some claims that are much higher than the average. (b) A random sample of 40 collision claims of 30- to 59-year-old drivers results in a mean claim of $3680 with a standard deviation of $2030. An independent random sample of 40 collision claims of 20- to 24-year-old drivers results in a mean claim of $4536 with a standard deviation of $2306. Using the concept of hypothesis testing, determine if a higher insurance premium should be paid by 20- to 24-year-old drivers. Use a a = 0.10 level of significance. Let population 1 be all collision claims of 30- to 59-year-old drivers and let population 2 be all collision claims of 20- to 24-year-old drivers. Write the hypotheses for the test. Ho: Hy H2 H1: H1 Calculate the test statistic. to = -1.76 (Round to two decimal places as needed.) Identify the critical value(s). Select the correct choice below and fill in the answer box(es) within your choice. (Round to two decimal places as needed.) O A. F1-a/2, n, - 1, n2 -1 = and Fa/ 2, n, - 1, n2 - 1= O B. to = - 1.30 O c. -ta/2 = and ta/2= O D. OE. -Za/2 = and Za/2= OF. Fa, n, - 1, n2 - 1= O G. F1-a, ny - 1, 02 -1= What is the conclusion? Reject the null hypothesis and conclude there is sufficient evidence that the mean claim of 30- to 59-year-old drivers is less than the mean claim of 20- to 24-year-old drivers at the o = 0.10 level of significance. It seems that 20- to 24-year-old drivers should pay a higher insurance premium. Help Me Solve This View an Example Get More Help - Similar Question2 Automobile collision insurance is used to pay for any claims made against the driver in the event of an accident. This type of insurance will typically pay to repair any assets that your vehicle damages. Complete parts (a) and (b). Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Click here to view the table of critical F-values (page 1). Click here to view the table of critical F-values (page 2). Click here to view the table of critical F-values (page 3). Click here to view the table of critical F-values (page 4). (a) Collision claims tend to be skewed right. Why do you think this is the case? O A. Most claims are either low or high value, with not many claims for moderate values. O B. Most claims are high, but there are some claims that are much lower than the average. O C. Most claims are mode ome claims that are much higher than the average. O D. Most claims are moderate with similar numbers of low value and high value claims. X Table of Critical t-Values F-Distribution Critical Values Table (page 1) F-Distribution Critical Values Degrees of Freedom in the Numerator Area in Area in right tad Right Tail 2 8 Right Tail 1-Distribution Standard Normal Distribution Table (page 1) 0.100 39.86 49.59 53.59 55.83 57.24 58.20 58.91 59.44 0.100 Area in Right Tail 0.050 161.45 199.50 215.71 30.16 64779 799.50 864.16 899.58 921.85 233.99 36.77 0.025 93711 948.22 238.88 0. Degrees of 956.66 0. Freedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 0.01 0.005 0.0025 0.001 0.0605 0.010 5763.65 5858.99 0:001 4052.20 4999.50 5403.35 5624.58 5928.36 5981.07 0.010 05284.07 499999.50 40379.20 62499.58 76404.56 58593711 592873.29 98144.16 0.001 12732 1 22.327 636.619 0.100 9.00 9.16 9.24 9.29 9.33 9.35 937 0.100 0.050 18.51 19.00 19.16 19.25 19.30 19.33 19.35 19.37 0.050 0.025 38.51 39.00 39.17 39.25 39.30 39.33 39.36 0.025 0.010 98.50 99.00 99.17 99.25 99.30 99.33 99.36 0.010 0.001 098.50 299.00 999.17 999.25 990 30 999.33 999.36 0.001 0.100 5.54 5.46 5.39 5.34 5.31 5.28 5.27 5.25 0.100 0.050 10.13 9.55 9.28 9.12 9.01 8.94 8.89 8.85 0.050 Standard Normal Distribution 3 0.025 17.44 16.04 15.44 15.10 14.73 14.62 0.025 2767 0.010 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.010 34.12 30.82 29.46 28.71 28.24 0:901 67.03 148.50 141.11 13710 134.58 132.85 31.58 130.62 0.001 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 0.100 4.32 3.98 0.100 6.59 0.050 0.0905 0:0907 0.050 6.09 0,9098 9.07 0.6013 0.0011 0.0010 0.0010 0.025 10.65 9.98 9.60 9.20 $0.025 0.0014 0.010 21.20 18.00 16.69 15.98 15.52 15.21 14.98 0.010 0.001 74 14 61.25 56.18 53.44 51.71 50.53 49.66 49.00 0.001 0.0020 Degrees of Freedom in the Denominator Degrees of Freedom in the Denominator 0.100 4.06 3.78 3.62 3.52 3.45 3.40 3.37 3.34 0.100 0.050 6.61 5.79 5.41 5.05 4.88 4.82 0.050 5 0.025 10.01 8.43 7.15 6.85 0.025 S 0.010 16.26 13.27 12.06 11.39 10.97 10.67 10.46 0.010 0901 A718 37.12 33.20 31.09 29.75 28.83 28.16 2765 0.001 0.0197 0.100 3.46 3.29 3.18 3.01 298 0.10 0.025 0.050 5.99 4.53 4.21 4.15 0.050 5.99 5.60 0.0314 0.025 8.81 6.60 6.23 0.010 13.75 10.92 9.78 9.15 8.75 8:26 0.010 0.001 35.51 2700 23.70 21.92 20.80 20.03 19.46 19.03 0.001 0.100 3.59 3.26 3.07 2.96 2.88 2.83 2.78 2.75 0.100 0.050 5.59 4.35 4.12 3.97 3.87 3.79 3.73 0.050 0.025 8.07 0.025 0.010 12.25 9.55 8.45 719 6.99 0.010 0:001 21.69 18.77 1720 16.21 15.52 15.02 14.63 0.001 0.100 3.46 3.11 2.92 2.81 2.73 2.67 2.62 2.59 0.100 0.050 5.32 4.07 3.84 3.69 3.58 3.50 0.050 0.025 6.06 5.42 5.05 4.82 4.65 4.53 0.025 0.010 11.26 8.65 6.37 6.03 0.010 0.001 25.41 18.49 15.83 14.39 13:48 12.86 12.40 12.05 0.001 0.20 0.10 0.001 0.0605 Area in Area in Degrees of 0.25 0.05 0.025 0.02 0.01 0.005 0.9025 Right Tail Freedom 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 Right Tail Degrees of Freedom in the Numerator -Distribution 0.06 0.09 Area in Right Tail3, sample The data table represents the measure of a variable before and after a treatment. Does the sample evidence suggest that the treatment is effective in decreasing the value of the response variable? Use the a = 0.10 level of significance. Complete parts (a) through (d). Individual 1 Before, Xi 42 46 After, yi 37 48 (a) What type of test should be used? Choose the correct answer below. O A. A hypothesis test regarding two population standard deviations. B. A hypothesis test regarding the difference between two population proportions from independent samples. O C. A hypothesis test regarding the difference of two means using Welch's approximate t. D. A hypothesis test regarding the difference of two means using a matched-pairs design. ne the null and alternative hypotheses. Let Hd = Hx - Hy. Choose the correct answer below. OA. Ho: Hd # 0; H1: Hd > 0 B. Ho: Hd = 0; H,: Hd >0 O C. Ho: Hd = 0; H1 : Hd # 0 OD. Ho: Hd = 0; H,: Hd a. X B. There is sufficient evidence to reject the null hypothesis because the P-value a. Help Me Solve This View an Example Get More Help - Similar QuestionThe data table represents the measure of a variable before and after a treatment. Does the sample evidence suggest that the treatment is effective in decreasing the value of the response variable? Use the a = 0.10 level of significance. Complete parts (a) through (d). Individual 1 Before, Xi 38 After, yi 39 33 (a) What type of test should be used? Choose the correct answer below. O A. A hypothesis test regarding the difference between two population proportions from independent samples. O B. A hypothesis test regarding two population standard deviations. O C. A hypothesis test regarding the difference of two means using a matched-pairs design. O D. A hypothesis test regarding the difference of two means using Welch's approximate t. Help Me Solve This View an Example Get More Help - Clear All Check Answer4, sample Use the following information to complete steps (a) through (d) below. A random sample of n, = 135 individuals results in x1 = 40 successes. An independent sample of n2 = 150 individuals results in X2 = 60 successes. Does this represent sufficient evidence to conclude that p, P2: H1 : P1 P2 O C. Ho: P1 = P2: H1 : P1 # P2 D. HO: P1 = P2: H1: P1 a. B. There is not sufficient evidence to reject the null hypothesis because the P-value a. D. There is sufficient evidence to reject the null hypothesis because the P-value H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value = 0.0037 State the researcher's conclusion. Which of the following is correct? O A. Fail to reject Ho, there is not sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women B. Fail to reject Ho, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women C. Reject Ho, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women O D. Reject Ho, there is not sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women (c) What is the 95% confidence interval for the mean difference in pulse rates of men versus women? The lower bound is - 10.91. The upper bound is - 1.69 (Round to two decimal places as needed.) Interpret this result. O A. 95% percent of the time the mean difference is in the confidence interval. B. 95% percent of the time the means are in the confidence interval. C. We are 95% confident that the mean difference is in the confidence interval. O D. We are 95% confident that the means are in the confidence interval. Help Me Solve This View an Example Get More Help - Similar Question6 A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 51 men and 78 women to participate in Two sample T for Men vs Women the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Mean StDev SE Mean Men 112.9 11.6 1.6 Women 18. 14.2 16 90% CI for mu Men - mu Women (-9.81, -2.19) T-Test mu Men = mu Women (vs H2 OB. HO: Hy = H2; Hai My # H2 O C. Ho: My = H2i Ha: My