SSTR = 6750 SSE = 8000 nt= 20 Ho: u1=u2=u3=u4 Ha: at least one mean is...
SSTR = 6750 SSE = 8000 nt= 20
Ho: u1=u2=u3=u4
Ha: at least one mean is different
The test statistic to test the null hypothesis equals:
4.5 0.84 0.22 4.22
Homework Answers
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SSTR = 6750 SSE = 8000 nt= 20
Ho: u1=u2=u3=u4
Ha: at least one mean is...
SSTR= 6750 SSE = 8000 nt = 20Ho: u1=u2=u3=u4 Ha: at least one mean is differnt...
SSTR= 6750 SSE = 8000 nt = 20Ho: u1=u2=u3=u4 Ha: at least one mean is differnt At alpha=.05 the null hypothesis A. was designed incorrectly B. None of these C. Should be rejected D. Should not be rejected
VW 03 DOLL SSTR =6750 SSE = 8000 nt=20 HouM,=M2 = M3=M4 Hai at least one...
VW 03 DOLL SSTR =6750 SSE = 8000 nt=20 HouM,=M2 = M3=M4 Hai at least one mean is different The test statistic to test the null hypothesis equals A. 4,22 B. 0,84 C. 0.22 D.4.5
D Question 18 SSTR-6.750 SSE - 4,000 n - 20 Ha Ha gult gulla at least...
D Question 18 SSTR-6.750 SSE - 4,000 n - 20 Ha Ha gult gulla at least one mean is different H The test statistic to test the null hypothesis equals: 0.22 0.84 6.00 4.5 1 pts D Question 19
Please answer this question based on the following data. SSTR = 6,750 H0: μ1= μ2=μ3=μ4 SSE =...
Please answer this question based on the following data. SSTR = 6,750 H0: μ1= μ2=μ3=μ4 SSE = 8,000 Ha: at least one mean is different Nτ = 20 The null hypothesis is to be tested at the 5% level of significance. The p-value is less than .01 between .01 and .025 between .025 and .05 between .05 and .10
A sample mean, sample standard deviation, and sample size are given. Use the one meant test...
A sample mean, sample standard deviation, and sample size are given. Use the one meant test to perform the required hypothesis test about the mean, , of the population from which the sample was drawn. Use the P-value approach. Also, assess the strength of the evidence against the null hypothesis. x-22,298, s=14200, n = 17, HO: P = 30,000, Ha# 30,000 a -0.05. Test statistic: 224. P.value 0.0200. Reject the null hypothesis. There is sufficient evidence to conclude that the...
The options for each of part c. are: - At least one mean is different -...
The options for each of part
c. are:
- At least one mean is different
- The means are the same.
- The standard deviations are different.
- The standard deviations are the same.
Exercise 12-15 (LO12-4) The following data are given for a two-factor ANOVA with two treatments and three blocks. Treatment Block 1 2 A 40 36 B 33 27 C 46 39 Using the 0.05 significance level conduct a test of hypothesis to determine whether the block...
Consider the following hypotheses: Ho: 4 = 150 HA: H < 150 A sample of 80...
Consider the following hypotheses: Ho: 4 = 150 HA: H < 150 A sample of 80 observations results in a sample mean of 144. The population standard deviation is known to be 28. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal...
Consider the following hypotheses: HO: > 220 HA: U <220 A sample of 72 observations results...
Consider the following hypotheses: HO: > 220 HA: U <220 A sample of 72 observations results in a sample mean of 209. The population standard deviation is known to be 18. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test...
In an experiment to compare the tensile strengths of I = 4 different types of copper...
In an experiment to compare the tensile strengths of I = 4 different types of copper wire, ) = 3 samples of each type were used. The average between group and average within group estimates were computed as MST = 2676.3 and MSE = 1190.2, respectively. Often hypothesis tests are conducted using what is referred to as the rejection region method. This compares the calculated F test statistic to an F value that cuts off a "rejection region". If the...
5.48 Work hours and education: The General Social Survey collects data on demographics, education, and work,...
5.48 Work hours and education: The General Social Survey collects data on demographics, education, and work, among many other characteristics of US residents. Using ANOVA, we can consider educational attainment levels for all 1,172 respondents at once. Below are the distributions of hours worked by educational attainment and relevant summary statistics that will be helpful in carrying out this analysis. Educational attainment Less than HS (1) HS (2) Jr Coll (3) Bachelor's (4) Graduate (5) Total Mean 38.67 39.6 41.39...
VW 03 DOLL SSTR =6750 SSE = 8000 nt=20 HouM,=M2 = M3=M4 Hai at least one mean is different The test statistic to test the null hypothesis equals A. 4,22 B. 0,84 C. 0.22 D.4.5
D Question 18 SSTR-6.750 SSE - 4,000 n - 20 Ha Ha gult gulla at least one mean is different H The test statistic to test the null hypothesis equals: 0.22 0.84 6.00 4.5 1 pts D Question 19
A sample mean, sample standard deviation, and sample size are given. Use the one meant test to perform the required hypothesis test about the mean, , of the population from which the sample was drawn. Use the P-value approach. Also, assess the strength of the evidence against the null hypothesis. x-22,298, s=14200, n = 17, HO: P = 30,000, Ha# 30,000 a -0.05. Test statistic: 224. P.value 0.0200. Reject the null hypothesis. There is sufficient evidence to conclude that the...
The options for each of part
c. are:
- At least one mean is different
- The means are the same.
- The standard deviations are different.
- The standard deviations are the same.
Exercise 12-15 (LO12-4) The following data are given for a two-factor ANOVA with two treatments and three blocks. Treatment Block 1 2 A 40 36 B 33 27 C 46 39 Using the 0.05 significance level conduct a test of hypothesis to determine whether the block...
Consider the following hypotheses: Ho: 4 = 150 HA: H < 150 A sample of 80 observations results in a sample mean of 144. The population standard deviation is known to be 28. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal...
Consider the following hypotheses: HO: > 220 HA: U <220 A sample of 72 observations results in a sample mean of 209. The population standard deviation is known to be 18. (You may find it useful to reference the appropriate table: z table or t table) a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test...
In an experiment to compare the tensile strengths of I = 4 different types of copper wire, ) = 3 samples of each type were used. The average between group and average within group estimates were computed as MST = 2676.3 and MSE = 1190.2, respectively. Often hypothesis tests are conducted using what is referred to as the rejection region method. This compares the calculated F test statistic to an F value that cuts off a "rejection region". If the...
5.48 Work hours and education: The General Social Survey collects data on demographics, education, and work, among many other characteristics of US residents. Using ANOVA, we can consider educational attainment levels for all 1,172 respondents at once. Below are the distributions of hours worked by educational attainment and relevant summary statistics that will be helpful in carrying out this analysis. Educational attainment Less than HS (1) HS (2) Jr Coll (3) Bachelor's (4) Graduate (5) Total Mean 38.67 39.6 41.39...
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