1. a)Researchers are comparing the number of events that occurred post-intervent

Question

1. a)Researchers are comparing the number of events that occurred post-intervention between two treatment groups Because many of the participants had no events and the distribution for the number of events was very skewed they tested this using a method called the negative binomial model. This method adjusts the estimated standard error to account for the non-normality. They found a model of the form (# events) = 0.49 + 0.52*group. The interpretation and testing of coefficients here is similar to that of linear regression. If the two treatments groups are coded as 1=usual care and 0=new treatment, what is the difference in the mean number of events between the two groups?

b)Using a normal approximation for the test statistic, if the reported standard error for the slope was 0.30 what is the p-value for testing the above? (Approach the same way you would a linear regression model.)

c)What do you conclude for the above about the mean number of events between the two groups? Justify your answer.

Explanation / Answer

for new treatment.

mean=0.49+0.52*0=0.49

for usual care

mean=0.49+0.52*1=1.01

difference of mean=1.01-0.49=0.52 ( which is same as slope)

(b) t=slope/SE(slope)=0.52/0.3=1.733

since number of observation is to mentioned, so it is difficult to calculate p-value.

using normal approximation of t-distribution

the two tailed p-value=0.084 ( using normsdist(1.733) ms-excel command)

(c) at 5% level of significance both group donot differ as p-value is more than alpha=0.05

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