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

Section: Chapter Questions

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

See similar textbooks

Similar questions

The number of semester hours of college credit taken by first-time candidates for a certain professional exam has a Normal distribution with a mean of 120 hours and a standard deviation of 16 hours. Consider a random sample of n first-time candidates for the exam and let x represent the mean number of hours of college credit taken for the sample. Complete parts a through e below. Click the icon to view the table of normal curve areas. ..... a. Describe the shape of the sampling distribution of x. A. The shape is Normal distribution only when sample size is greater than 30 B. The shape is Normal distribution for any sample size O C. is always the same as the distribution of the population D. The shape is skewed right b. If 16 first-time candidates for the exam are randomly selected, what is the probability that the mean number of hours of college credit taken by these students x is between 118 and 122 hours? P(118 < x <122) = (Round to four decimal places as needed.) c. If 64 first-time…
Test Nondefective 79 93 81 93 79 94 79 91 76 94 81 94 79 94 79 92 82 93 86 93 83 92 78 95 77 92 77 93 80 95 73 95 87 95 75 93 82 96 74 93 81 95 81 95 82 94 73 94 75 94 82 92 87 95 84 94 86 94 79 94 82 95 70 92 86 96 84 94 75 95 75 95 77 93 78 95 76 95 84 94 76 94 78 94 81 94 79 94 78 94 above is excel data and please solve the excel part with formulas and explanation, thanks
7.
Number of ChildrenTable 1 shows the number of women (per 1000) between 15 and 50 years of age who have been married grouped by the number of children they have had. Table 2 gives the same information for women who have never been married.1 Number of Children Women per 1000 0 162 1 213 2 344 3 182 4 69 5+ 32 Table 1 Women who have been married Number of Children Women per 1000 0 752 1 117 2 72 3 35 4 15 5+ 10 Table 2 Women who have never been married 1http://www.census.gov/hhes/fertility/data/cps/2014.html, Table 1, June 2014. Condition gt requires numbers instead of "null" and "0" (a) Without doing any calculations, which of the two samples appears to have the highest mean number of children? Women who have been marriedWomen who have never been married Which of the distributions appears to have the mean most different from the median? Women who have been marriedWomen who have never been married…
QUESTION 6 Every day, Kim empties the change out of her pockets and stores it in a large jar for a Rainy Day Fund. She keeps track of how much change she empties from her pockets each day over the span of several months and stores the results in the five number summary below. Min: 35¢ Q1: 75¢ Med: 90¢ Q3: 95¢ Max: 105¢ Select the boxplot that represents this data. Kim's Rainy Day Fund 30 35 40 15 50 55 60 65 70 75 80 85 90 95 100 105 110 cents Kim's Rainy Day Fund 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 cents
please help me solve this practice question using excel functions 7065 6870 6858 6641 6306 6671 6663 6492 6569 6601 6653 6729 7237 6911 6784 7312 7313 7082 7462 7311 6994 7399 7231 7666 7574 7706 7561 7198 7255 7116 7362 7511 7773 7776 7547 7567 7679 7627 7883 7274 7072 7165 7025 7125 7195 7318 7461 7624 7717 7450 7108 7053 6768 6896 7158 7152 7171 6902 6968 6619 6427 6567 6422 6686 6976 7069 7222 7381 7495 7460 7721 7804 7558 7781 7517 7795 7976 8038 8122 7854 7924 7754 7824 7935 8233 8304 8094 7974 7659 7538 7282 7435 7399 7712 8259 7935 8142 7696 7366 7182 7184 7408 7408 7805 7187 7074 7141 7245 7518 7748 7620 7434 7489 7345 7557 7490 7625 7691 7698 8019 8015 8064 8068 7839 7923 7939 8215 8730 8487 8444 8364 7893 8183…
The accompanying data are 45 commute times to work in minutes for workers of age 16 or older in Chicago. Construct a frequency distribution. Use a class width of 15 minutes and begin with a lower class limit of 0 minutes. Do the data amounts appear to have a normal distribution? Examine the data and identify anything appearing to be unique. Click the icon to view the commute times. Frequency Time (minutes) 0-14 4 15-29 15 30-44 11 45-59 9 60-74 75-89 5 1 (Type whole numbers.) Do the data amounts appear to have a normal distribution? OA. No, because while the distribution is approximately symmetric, the frequencies start at a maximum and become low. B. No, because while the frequencies start low, proceed to one or two high frequencies, then decrease to a low frequency, the distribution is not symmetric. OC. Yes, because the frequencies start low, proceed to one or two high frequencies, then decrease to a low frequency, and the distribution is approximately symmetric. OD. No, because the…
75 70 65 60 55 55 55 51 50 45 40 35 59 63 60 62 66 43 56 56 60 69 73 64 46 50 70 75 70 63.2 65 55 60 60 52.75 55 51 50 47 Franklin Roosevelt Franklin Roosevelt Franklin Roosevelt Franklin Roosevelt Harry Truman Dwight Eisenhower Dwight Eisenhower John F. Kennedy Lyndon Johnson Richard Nixon Richard Nixon Jimmy Carter Ronald Reagan Ronald Reagan George H.W. Bush Bill Clinton Bill Clinton George W. Bush George W. Bush Barack Obama Barack Obama Donald Trump 1932 1936 1940 1944 1948 1952 1956 1960 1964 1963 1972 1975 1980 1984 1983 1992 1996 2000 2004 2008 2012 Democratic nominees who win Republican nominees who win 2016 1932 - 2016 If I wanted to test if there is a statistically significant difference between the average age of Democratic vs Republican presidents. Which test should I use? O One Sample t-test, one-tail O Independent Samples t-test, one tailed O Independent Samples t-test, two tailed O One Sample t-test, two-tail 45 40 35
12 11 12 10 12 11 12 11 10 10 10 1:11 12 11 12 11 12 11 12 11 10 10 10 10 6. 12 11 12 11 12 11 12 11 10 10 10 10 12. 1. 12 11 12 11 11 10 10 10 10 RI 8. 91 91 3. 1.
[112, 158, 184, 205, 222, 231, 266, 268, 269, 301, 347, 1026, 1694, 1741, 2031, 2381, 2389, 2660, 2704, 2834, 2876, 3150, 3238, 3289, 3387, 3517, 3563, 3577, 3621, 3701, 3732, 3757, 3775, 3870, 3935, 4034, 4096, 4134, 4193, 4243, 4473, 4509, 4511, 5049, 5438] please answer these questions using the above data

Recommended textbooks for you

A First Course in Probability (10th Edition)

Probability

ISBN:

9780134753119

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability

Probability

ISBN:

9780321794772

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability (10th Edition)

Probability

ISBN:

9780134753119

Author:

Sheldon Ross

Publisher:

PEARSON

A First Course in Probability

Probability

ISBN:

9780321794772

Author:

Sheldon Ross

Publisher:

PEARSON

SEE MORE TEXTBOOKS

GET THE APP

About FAQ Academic Integrity Sitemap Document Sitemap

Contact Bartleby Contact Research (Essays)High School Textbooks Literature Guides Concept Explainers by Subject Essay Help Mobile App

GET THE APP

Privacy

Your CA Privacy Rights

Your NV Privacy Rights

Cookie Policy

About Ads

Manage My Data

bartleby, a Learneo, Inc. business

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