Question

# students surveyed 100 random medical professionals The students were interested in testing if the average patient...

students surveyed 100 random medical professionals The students were interested in testing if the average patient ratings from all 100 medical professionals surveyed were different from 3.5 (last year’s average patient ratings). Below are summary statistics of the variable “patient ratings” for the 100 medical professionals surveyed Mean Median n St. Dev. Std. Error of the Mean 5.193 6.89 100 0.735 0.1195 What is the appropriate way to construct the corresponding p-value to test this hypothesis?

(a) 2normcdf(|23.03401|, 9999, 99)

(b) 2tcdf(−9999, | − 2.312589|, 99)

(c) 2tcdf(−9999, |23.03401|, 99)

(d) 2tcdf(| − 2.312589|, 9999, 99)

(e) 2tcdf(|23.03401|, 9999, 99)

Solution:

The correct option is (e) 2tcdf(|23.03401|, 9999, 99)

Explanation:

To use the calculator to find the p-value, we need the test statistic, and degrees of freedom.

In the given problem, the test statistic is:

The calculator function is:

Since the given test is two-tailed, so we have to add 2 in front of the function.

Lower = 23.03401,

Upper = Any large number like 9999

df = 99

Therefore, the correct function is:

(e) 2tcdf(|23.03401|, 9999, 99)

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