MUST USE MINITAB!!! A study was conducted to assess the effectiveness of a new a

Question

MUST USE MINITAB!!!

A study was conducted to assess the effectiveness of a new antibiotic treatment for strep throat in children. Children 6 to 14 who met the entry criteria were randomized to one of three treatment groups. Group 1 was given standard treatment 1, group 2 was given standard treatment 2, and the last group (group 3) was given the new antibiotic treatment. (The 2 standard treatments are different.) The response measured on each child was the number of days to cure the strep infection. Use the partial ANOVA table provided to answer the questions below. a. This output would be generated in order to test what set of hypotheses? Null: Alternative: Assume for the parts below that the ANOVA conditions are satisfied. a. Provide the missing value of the treatment df and the F test statistic. b. What is your best estimate of the common population variance assumed by the ANOVA? (Provide a numerical value.) c. Using a .05 significance level, what is your conclusion (in context) for this ANOVA?

Explanation / Answer

This data does not require MINITAB at all... since we are not calculating the ANOVA from the raw data

If raw data ( data values of different treatments) is given we can use, ANOVA one way directly

From the table given below,

The missing values are:

a) Null hypothesis: There is no difference between treatments

Alternate hypothsis: There is difference between treatments

b) Provided abive in the table

c) We have to find degrees of freedom

n1=2, n2=30

degrees of freedom (n-1,n2-1) at 95% CI,

At (2-1,30-1) = (1,29) degrees of freedom , F value = 4.1829 (critical value)

From F table

d) We got F value as per ANOVA table

F = 4.0734

F critical = 4.0734

Since F < F critical and p value = 0.27 > 0.05, we can conclude that there is no significance difference between treatments

hence, We do not reject null hypothesis

Source SS Df MS F p-value Treatment 38.364 (32-30)=2 19.182 19.182/4.709 = 4.0734 0.27 Residuals 141.273 30 4.709 Total 179.636 32

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