Question

Problem: The Striving for Perfection Urgent Care Center feels that its patients wait time is too long. The staff has noticed that increasing numbers of people are waiting longer in the waiting room. The national benchmark performance is 90 minutes or less from the time the customer arrives to the time the customer is treated and discharged.

The staff decides to examine the time it takes for all patients to move through their service process on two successive days (columns A & B of the data set are one sample).

Questions:

1a- Is the variable categorical, discrete or continuous?

1b- What type of sampling did they use?

1c- Summarize these data graphically and using data description techniques.

1d- Is the clinic meeting the national benchmark time? ((Test the hypothesis that the clinic's population mean is less than 90 minutes. Use a 95% confidence level. Assume the variable is normally distributed. State whether the assumptions for the test are met or not)

Find AaBbCcReplace 11 103 30 61 22 66 90 220 99 170 120 150 160 Copy 42-30 40 57 58 42 38 4% 75 4T 24 70 54 t, select, share Upload 49 50 55 90 45 te ST 30 55 22-50 58 56 55 56 47 A-之ーー···-·흐'L. d 65 Lat 190 200 90 97 95-61 125 49 hus 47 B 1 33-30 93 95 111 u-kx, x 24 22 00 90 220 99 250-120 150 'Norm al 態38 78 170 47 24 93 05 111 80 30 61 2-00 97 95-30 125 11 18 00 140 99 210 120 50 160 30 95 30 125 Heading i 70 Title Ti 'No Spac Heading 2 Subtitle Format Painter Select- F Editing Clipboard Font Paragraph Styles Column A Column D Celuma G Urgent Care facility patient imes in mple: Sampla Elapsed tines after prnees improved IYPE OF 166 SERVICCF 130 anged 178 178 146 245 243 243 124 124 143

Answer 1

1a) The variable is a continuous variable

b) The sampling type is population sampling as all the daily population is sampled here

c)

30 Compare Day wise 10 Day 1 summary Day 2 Summary Mor 30 507090 110 130 150170 190 210 230 250 270 Day 1 summary 54 3 6 65 61 3 1 0 3 1 0 Day 2 Summary 9 3 3 5 76 42 1 0 0 22 0

The above shows the summary of the data.

where mean of day 1 = 112.3 and standard deviation = 65.137  

where mean of day 2= 101.5 and standard deviation = 67.111

d) for day 1 we can see how the daily mean deviates from the standard 90 min.

assuming normal distribution Z = = 2.296

(X-a)/(o/v n) = (90-112.3)/(65, 137/V45)

so the P value = 0.011 so we can say that the clinic is not maintaining the standard for day1

for day 2 if we do the same analysis

assuming normal distribution Z = = 1.149

(X-a)/(o/v n) = (90-101.5)/(67, 111/V45)

so the p value is 0.125. So the clinic is matching the standard for day 2