Question

In: Statistics and Probability

Irritable bowel syndrome (IBS) is a nonspecific intestinal disorder characterized by abdominal pain and irregular bowel habits.

QUESTION 1

  1. Questions 1 through 4 refer to the following:

    Irritable bowel syndrome (IBS) is a nonspecific intestinal disorder characterized by abdominal pain and irregular bowel habits. Each person in a random sample of 24 patients having periodic attacks of IBS was randomly assigned to one of three treatment groups, A, B, and C. The number of hours of relief while on therapy is recorded for each patient in the table below.

    Treatment

    A

    B

    C

    11.2

    11.6

    23.1

    8.0

    12.5

    9.9

    3.1

    10.6

    12.1

    10.9

    11.6

    13.0

    12.3

    13.9

    13.3

    9.9

    8.6

    11.0

    9.1

    18.9

    15.2

    5.0

    19.1

    15.9

    sample mean

    8.6875

    13.35

    14.1875

    sample variance

    10.1698

    14.4486

    16.9270

    Give the value of the analysis of variance test statistic used to test H0: μA = μB = μC.

1 points   

QUESTION 2

  1. What's the critical value if the .05 significance level is used?

1 points   

QUESTION 3

  1. What's the p-value?

1 points   

QUESTION 4

  1. What's the decision using the .05 significance level?

    Reject H0

    Don't Reject H0

    Reject HA

    Don't Reject HA

Solutions

Expert Solution

Question 1

Grand mean, = (8.6875 + 13.35 + 14.1875) / 3 = 12.075

Treatment Sum of Squares (SSTR) =   where rj is the number of rows in the jth treatment

= [ 8 *(8.6875 - 12.075)2 + 8 *(13.35 - 12.075)2 + 8 *(14.1875 - 12.075)2 ]

= 140.5075

Error Sum of Squares (SSE) = (rj - 1) sample variances

= (8 - 1) * 10.1698 + (8 - 1) * 14.4486 + (8 - 1) * 16.9270

= 290.8178

Degree of freedom for treatments, dftr = Number of treatments - 1 = 3 - 1 = 2

Degree of freedom for Error, dfe = Number of observations - Number of treatments = 8 * 3 - 3 = 21

MSTR = SSTR / dftr = 140.5075 / 2 = 70.25375

MSE = SSE / dfe = 290.8178 / 21 = 13.84847

Test statistic, F = MSTR / MSE = 70.25375 / 13.84847 = 5.07

QUESTION 2

Critical value of F for the .05 significance level and df = 2, 21 is 3.47

QUESTION 3

P-value = P(F > 5.07, df = 2, 21) = 0.016

QUESTION 4

Since p-value is less than 0.05 significance level, we reject null hypothesis H0  

Reject H0

orchestra answered 2 years ago
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