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f. Conduct a two-tailed comparison of the two groups with a=.05. g. Now combine the observation from the tables mentioned in parts (a) and (d). so that are 18 observations for each Treatment. What is the s for treatment 1 and treatment 2? What is M?

f. Conduct a two-tailed comparison of the two groups with a=.05. g. Now combine the observation from the tables mentioned in parts (a) and (d). so that are 18 observations for each Treatment. What is the s for treatment 1 and treatment 2? What is M?

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
Section10.4: Distributions Of Data
Problem 19PFA
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Looking for a detailed answer to the sub parts g and f please. Thanks.

graph Styles 6. Check out the frequency distributions for two treatment groups: Treatment 1I Treatment 2 n = 9 M= 8 n = 9 M= 13 S = 1.22 S = 1.22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 a. Write a table with observations for these samples. b. Using the table, explain (calculate) where the S=1.22 comes from C. Conduct a two-tailed comparison of the two groups with a=.05. Please remember to state your hypothesis. Now look at: Treatment 1I Treatment 2 O n = 9 M= 8 S = 6.65 n = 9 M= 13 S = 6.65 2 3 5 6 7 8 9 10 11 12 13 14 15 l6 17 18 19 20 21 d. Write a table with observations for these samples. e. Using the table, explain (calculate) where the S=6.65 comes from Conduct a two-tailed comparison of the two groups with a=.05. f. g. Now combine the observation from the tables mentioned in parts (a) and (d), so that are 18 observations for each Treatment. What is the s for treatment 1 and treatment 2? What is M? h. Conduct a two-tailed comparison of the two groups with a-.05.
Transcribed Image Text:graph Styles 6. Check out the frequency distributions for two treatment groups: Treatment 1I Treatment 2 n = 9 M= 8 n = 9 M= 13 S = 1.22 S = 1.22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 a. Write a table with observations for these samples. b. Using the table, explain (calculate) where the S=1.22 comes from C. Conduct a two-tailed comparison of the two groups with a=.05. Please remember to state your hypothesis. Now look at: Treatment 1I Treatment 2 O n = 9 M= 8 S = 6.65 n = 9 M= 13 S = 6.65 2 3 5 6 7 8 9 10 11 12 13 14 15 l6 17 18 19 20 21 d. Write a table with observations for these samples. e. Using the table, explain (calculate) where the S=6.65 comes from Conduct a two-tailed comparison of the two groups with a=.05. f. g. Now combine the observation from the tables mentioned in parts (a) and (d), so that are 18 observations for each Treatment. What is the s for treatment 1 and treatment 2? What is M? h. Conduct a two-tailed comparison of the two groups with a-.05.
Expert Solution
Step 1

f. For the two-tailed comparison for two groups, the null and alternative hypothesis are given by :

Ho : μ1= μ2H1 : μ1≠ μ2

The test statistics is given by Students' t-distribution because population standard deviation is unknown.

t=x1-x2s12n1+s22n2~tn1+n2-2

t=8-136.6529+6.6529=-53.135=-1.595

The two-tailed critical value for t-distribution at α=0.05 with (18-2)=16 degrees of freedom is ±2.120

The null hypothesis is rejected according to the decision rule when calculated test statistics is greater than the tabulated value and is accepted otherwise.

Since -1.595 < -2.120, null hypothesis is accepted concluding that the average for the two groups are same.

 

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