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11. The following experiment was conducted to compare two coatings designed to improve the durability of the soles of jogging shoes. A 1/8-inch layer of coating was applied to one of the pair of shoes and a layer of equal thickness of coating 2 was applied to the other shoe. Ten joggers were given pairs of shoes treated in this manner and were instructed to record the number of miles covered in each shoe before the 1/8-inch was worn through in any one place. The results are listed in the table: JOGGER 1 2 3 4 567 8 9 10 COATING 1 892 904 775 435 946 853 780 695 825 750 COATING 2 985 953 775 510 895 875 895 725 858 812

11. The following experiment was conducted to compare two coatings designed to improve the durability of the soles of jogging shoes. A 1/8-inch layer of coating was applied to one of the pair of shoes and a layer of equal thickness of coating 2 was applied to the other shoe. Ten joggers were given pairs of shoes treated in this manner and were instructed to record the number of miles covered in each shoe before the 1/8-inch was worn through in any one place. The results are listed in the table: JOGGER 1 2 3 4 567 8 9 10 COATING 1 892 904 775 435 946 853 780 695 825 750 COATING 2 985 953 775 510 895 875 895 725 858 812

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section: Chapter Questions
Problem 25SGR
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To the tutor who answered this please send me the formula for this and how to this on excel thank you.
11. The following experiment was conducted to compare two coatings designed to improve the durability of the soles of jogging shoes. A 1/8-inch layer of coating was applied to one of the pair of shoes and a layer of equal thickness of coating 2 was applied to the other shoe. Ten joggers were given pairs of shoes treated in this manner and were instructed to record the number of miles covered in each shoe before the 1/8-inch was worn through in any one place. The results are listed in the table: JOGGER 1 2 3 4 5 6 7 8 9 10 COATING 1 892 904 775 435 946 853 780 695 825 750 COATING 2 985 953 775 510 895 875 895 725 858 812
Transcribed Image Text:11. The following experiment was conducted to compare two coatings designed to improve the durability of the soles of jogging shoes. A 1/8-inch layer of coating was applied to one of the pair of shoes and a layer of equal thickness of coating 2 was applied to the other shoe. Ten joggers were given pairs of shoes treated in this manner and were instructed to record the number of miles covered in each shoe before the 1/8-inch was worn through in any one place. The results are listed in the table: JOGGER 1 2 3 4 5 6 7 8 9 10 COATING 1 892 904 775 435 946 853 780 695 825 750 COATING 2 985 953 775 510 895 875 895 725 858 812
12:41 Excel output is given below: t-Test: Paired Two Sample for Means Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Variable 1 Variable 2 785.5 828.3 18573.12222 21078.05556 10 0.944442232 f) The value of the test statistic is 2.836 ? 0 2.835909257 0.009766558 1.833112933 0.019533115 2.262157163 9 (rounded to 3 decimal places). The p-value is 0.019533 i.e., 0.020 (rounded to 3 decimal places). g) From the output, the critical value is 2.262 (two-tails). 10 Rejection region is the area beyond ±2.262. That is, reject Ho ift < -2.262 ort > 2.262 Using P-value method, the rejection rule is: Reject Ho, if p-value is less than or equal to the level of significance √x DO
Transcribed Image Text:12:41 Excel output is given below: t-Test: Paired Two Sample for Means Mean Variance Observations Pearson Correlation Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail Variable 1 Variable 2 785.5 828.3 18573.12222 21078.05556 10 0.944442232 f) The value of the test statistic is 2.836 ? 0 2.835909257 0.009766558 1.833112933 0.019533115 2.262157163 9 (rounded to 3 decimal places). The p-value is 0.019533 i.e., 0.020 (rounded to 3 decimal places). g) From the output, the critical value is 2.262 (two-tails). 10 Rejection region is the area beyond ±2.262. That is, reject Ho ift < -2.262 ort > 2.262 Using P-value method, the rejection rule is: Reject Ho, if p-value is less than or equal to the level of significance √x DO
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