ID Old ScoreNew Score 181822727237774474725787867772783848908697069107072117269127169137167147164157168167169177171187169197172207170217171227171237270247269257272267273277274287269297273307269317371327374337372347375357373367369377372387467397468407475417474427475437474447468457470467570477572487570497569507574517571527571537571547571557569567675577670587670597678607672617778627773637774647778657878667871677872687972697976707977717977728082738074748181758282Question 13 with the data set above. Page <12. Large Data Set 12 lists the scores on one round for 75 randomly selected members at a golf course, firstusing their own original clubs, then two months later after using new clubs with an experimental design.Denote the population of all golfers using their own original clubs as Population 1 and the population ofall golfers using the new style clubs as Population 2.http://www.flatworldknowledge.com/sites/all/files/data12.xlsa. Compute the 75 differences in the order original clubs- new clubs, their mean d, and theirsample standard deviation sd.b. Give a point estimate for ud = µ1 - H2 the difference in the mean score of all golfers using theiroriginal clubs and the mean score of all golfers using the new kind of clubs.c. Construct a 90% confidence interval for ud%3D9:3d. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at leastone stroke by using the new kind of clubs.13. JConsider the previous problem again. Since the data set is so large, it is reasonable to use the standardnormal distribution instead of Student's t-distribution with 74 degrees of freedom.a. Construct a 90% confidence interval for ud using the standard normal distribution, meaning that theformula is d ±Za/2 vn(The computations done in part (a) of the previous problem still apply andneed not be redone.) How does the result obtained here compare to the result obtained in part (c) ofthe previous problem?b. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at least onestroke by using the new kind of clubs, using the standard normal distribution. (All the work done inpart (d) of the previous problem applies, except the critical value is now z instead of t. (or the pvalue can be computed exactly instead of only approximated, if you used the p-value approach).) Howdoes the result obtained here compare to the result obtained in part (c) of the previous problem?C. Construct the 99% confidence intervals for Hd using both the t and z distributions. How muchdifference is there in the results now?

Use the given data to form the excel table: ID Old Score New Score d d-d¯2 1 81 82 -1 9.81777778 2 72 72 0 4.55111111 3 77 74 3 0.75111111 4 74 72 2 0.01777778 5 78 78 0 4.55111111 6 77 72 5 8.21777778 7 83 84 -1 9.81777778 8 90 86 4 3.48444444 9 70 69 1 1.28444444 10 70 72 -2 17.0844444 11 72 69 3 0.75111111 12 71 69 2 0.01777778 13 71 67 4 3.48444444 14 71 64 7 23.6844444 15 71 68 3 0.75111111 16 71 69 2 0.01777778 17 71 71 0 4.55111111 18 71 69 2 0.01777778 19 71 72 -1 9.81777778 20 71 70 1 1.28444444 21 71 71 0 4.55111111 22 71 71 0 4.55111111 23 72 70 2 0.01777778 24 72 69 3 0.75111111 25 72 72 0 4.55111111 26 72 73 -1 9.81777778 27 72 74 -2 17.0844444 28 72 69 3 0.75111111 29 72 73 -1 9.81777778 30 72 69 3 0.75111111 31 73 71 2 0.01777778 32 73 74 -1 9.81777778 33 73 72 1 1.28444444 34 73 75 -2 17.0844444 35 73 73 0 4.55111111 36 73 69 4 3.48444444 37 73 72 1 1.28444444 38 74 67 7 23.6844444 39 74 68 6 14.9511111 40 74 75 -1 9.81777778 41 74 74 0 4.55111111 42 74 75 -1 9.81777778 43 74 74 0 4.55111111 44 74 68 6 14.9511111 45 74 70 4 3.48444444 46 75 70 5 8.21777778 47 75 72 3 0.75111111 48 75 70 5 8.21777778 49 75 69 6 14.9511111 50 75 74 1 1.28444444 51 75 71 4 3.48444444 52 75 71 4 3.48444444 53 75 71 4 3.48444444 54 75 71 4 3.48444444 55 75 69 6 14.9511111 56 76 75 1 1.28444444 57 76 70 6 14.9511111 58 76 70 6 14.9511111 59 76 78 -2 17.0844444 60 76 72 4 3.48444444 61 77 78 -1 9.81777778 62 77 73 4 3.48444444 63 77 74 3 0.75111111 64 77 78 -1 9.81777778 65 78 78 0 4.55111111 66 78 71 7 23.6844444 67 78 72 6 14.9511111 68 79 72 7 23.6844444 69 79 76 3 0.75111111 70 79 77 2 0.01777778 71 79 77 2 0.01777778 72 80 82 -2 17.0844444 73 80 74 6 14.9511111 74 81 81 0 4.55111111 75 82 82 0 4.55111111 Sum 5615 5455 160 526.666667 Average 74.8666667 72.73333333 2.133333 d¯=∑dn=16075=2.1333s=∑(d-d¯)2n-1=526.666775-1=2.667792564≈2.6678S.E=sn=2.667875=0.308050151≈0.3081 a) Use the excel formula, "=NORM.S.INV(0.95)" for determining the critical value:zCritical = 1.645 C.I=x¯d±zα2*sdn=2.1333±1.645*2.667875=2.1333±0.5067C.I=(1.6266,2.6400) 90% confidence interval for μd using the standard normal distribution is C.I = (1.6266, 2.6400).b) Null hypothesis: H0: μD≥1 Alternate hypothesis: HA: μD<1 This is one tailed test. The significance level is 0.01 and we will conduct a matched-pairs t-test of the null hypothesis. Use the sample data to calculate the standard deviation of the differences(s), the standard error (S.E) of the mean difference, the degrees of freedom(DF), and the t statistic test statistic(t). z=[(x1-x2)-D]S.E=2.1333-10.3081=3.678351185≈3.68 The P-value for this test is the probability that a z statistic is less than 3.68. P-value = P(z < 3.68)Use the excel formula, "=NORM.S.DIST(3.68,TRUE)" to determine the P-value. P-value = 0.9999 Decision: Since the P-value (0.9999) is greater than the significance level (0.01), hence we failed to reject the null hypothesis. Conclusion: From the above test we have sufficient evidence in the favour of the claim that the mean golf score decreases by at least one stroke by using the new kind of clubs. c) Use the excel formula, "=NORM.S.INV(0.995)" for determining the critical value:zCritical = 2.576 C.I=x¯d±zα2*sdn=2.1333±2.576*2.667875=2.1333±0.7935C.I=(1.3398,2.9268) 99% confidence interval for μd using the standard normal distribution is C.I = (1.3398, 2.9268).

Math

Probability

one stroke by using the new kind of 13.Consider the previous problem again. Since the data set is so large, it is reasonable to use the standard normal distribution instead of Student's t-distribution with 74 degrees of freedom. a. Construct a 90% confidence interval for ud using the standard normal distribution, meaning that the formula is d ± Za/2 vn (The computations done in part (a) of the previous problem still apply and need not be redone.) How does the result obtained here compare to the result obtained in part (c) of the previous problem? b. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at least one stroke by using the new kind of clubs, using the standard normal distribution. (All the work done in part (d) of the previous problem applies, except the critical value is now Za instead of t, (or the p value can be computed exactly instead of only approximated, if you used the p-value approach).) How does the result obtained here compare to the result obtained in part (c) of the previous problem? C. Construct the 99% confidence intervals for ud using both the t and z distributions. How much difference is there in the results now?

one stroke by using the new kind of 13.Consider the previous problem again. Since the data set is so large, it is reasonable to use the standard normal distribution instead of Student's t-distribution with 74 degrees of freedom. a. Construct a 90% confidence interval for ud using the standard normal distribution, meaning that the formula is d ± Za/2 vn (The computations done in part (a) of the previous problem still apply and need not be redone.) How does the result obtained here compare to the result obtained in part (c) of the previous problem? b. Test, at the 1% level of significance, the hypothesis that the mean golf score decreases by at least one stroke by using the new kind of clubs, using the standard normal distribution. (All the work done in part (d) of the previous problem applies, except the critical value is now Za instead of t, (or the p value can be computed exactly instead of only approximated, if you used the p-value approach).) How does the result obtained here compare to the result obtained in part (c) of the previous problem? C. Construct the 99% confidence intervals for ud using both the t and z distributions. How much difference is there in the results now?

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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ID	Old Score	New Score
1	81	82
2	72	72
3	77	74
4	74	72
5	78	78
6	77	72
7	83	84
8	90	86
9	70	69
10	70	72
11	72	69
12	71	69
13	71	67
14	71	64
15	71	68
16	71	69
17	71	71
18	71	69
19	71	72
20	71	70
21	71	71
22	71	71
23	72	70
24	72	69
25	72	72
26	72	73
27	72	74
28	72	69
29	72	73
30	72	69
31	73	71
32	73	74
33	73	72
34	73	75
35	73	73
36	73	69
37	73	72
38	74	67
39	74	68
40	74	75
41	74	74
42	74	75
43	74	74
44	74	68
45	74	70
46	75	70
47	75	72
48	75	70
49	75	69
50	75	74
51	75	71
52	75	71
53	75	71
54	75	71
55	75	69
56	76	75
57	76	70
58	76	70
59	76	78
60	76	72
61	77	78
62	77	73
63	77	74
64	77	78
65	78	78
66	78	71
67	78	72
68	79	72
69	79	76
70	79	77
71	79	77
72	80	82
73	80	74
74	81	81
75	82	82