Question

Recent studies of the private practices of physicians suggested that the median and mean length of each patient visit was 25 minutes. A random sample of 17 visits to physicians in a certain private practice produced the visit lengths provided in the Excel spreadsheet. A researcher wants to use this data to determine if the median visit length of patients in this certain private practice is less than the recent study’s suggested median of 25 minutes. Using Minitab or other statistical software: 1. Conduct a sign test to answer the researcher’s question. Include relevant output from Minitab and use a significance level of α = 0.05. 2. How would your conclusion have changed if you used a significance level of α = 0.10? What does this say about your conclusion reached in part 1? 3. This test was to determine whether or not the median visit time was less than the time in the study. Do you think that the use of the median was appropriate in this case for the sample data set or could you have used a z-test for the mean? Answer this question by preparing a box plot and assessing the normality of the data set and explaining your conclusion.

Length of Visit
16.2
18.1
18.7
18.9
19.1
19.3
20.1
20.4
21.6
21.9
23.4
23.5
25.5
26.9
27.4
38.1
37.2

Answer 1

a) Sign test results from minitab

Sign Test for Median: Length of Visit

Method

η: median of Length of Visit

Descriptive Statistics

Sample	N	Median
Length of Visit	17	21.6

Test

Null hypothesis	H₀: η = 25
Alternative hypothesis	H₁: η < 25

Sample	Number < 25	Number = 25	Number > 25	P-Value
Length of Visit	12	0	5	0.072

given level of significance = 0.05, p-value obtained for this test is 0.072 which is greater than α, we fail to reject null hypothesis and we conclude that there is no significant evidence that median is less than 25 at 95 % confidence.

b) given α = 0.1, p-value is 0.072 which is less than α hence we reject null hypothesis and there is a significant evidence that to conclude median is less than 25 at 90% confidence.

3)

Box plot

Statistics

Variable	Skewness
Length of Visit	1.50

The box plot consists of two outliers and also the skewness is 1.5, which is heavily left skewed hence it cannot be approximated to normal distribution and z-test cannot be done.