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Q.13.2

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Introductory Statistics
Found in: Page 752
Introductory Statistics

Introductory Statistics

Book edition OER 2018
Author(s) Barbara Illowsky, Susan Dean
Pages 902 pages
ISBN 9781938168208

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Short Answer

MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 13.5.

Conc=0.6Conc=0.8Conc=1.0Conc=1.2Conc=1.4
916223027
6693147199168
9882120148132

Plot of the data for the different concentrations:

Test whether the mean number of colonies are the same or are different. Construct the ANOVA table (by hand or by using a TI-83,83+,or 84 calculator), find the p-value, and state your conclusion. Use a 5% significance level.

According to the output, the P-value is .79. So, at the 5% level of significance null hypothesis will be accepted. Thus, mean number of bacterial infections in hospitals were same for the various colony counts from different patients.

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Step by Step Solution

Step-1 Given Information

MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table shows various colony counts from different patients who may or may not have MRSA :

Conc=0.6Conc=0.8Conc=1.0Conc=1.2Conc=1.4
916223027
6693147199168
9882120148132

Step-2 Explanation

Consider the following Data of various colony counts from different patients to check the mean number of bacterial infections in hospitals were same

Conc=0.6Conc=0.8Conc=1.0Conc=1.2Conc=1.4
916223027
6693147199168
9882120148132

To check the mean number of number of bacterial infections in hospitals were same, use Ti-83calculator. For this, click on STAT press1EDIT, then put data into the listL1,L2,L3,L4,L5. The screenshot is given as below:

Now Again press STAT arrow over the TESTS arrow down to ANOVA Press var then select1, press var then select 2, press var then select 3, press var then select 4 and press var then select 5 to fill the values of L1,L2,L3,L4,L5. The screen shot is given as below:

Press ENTER the screenshot of output is given below:

According to the output, the P-value is .79. So, at the 5% level of significance null hypothesis will be accepted. Thus, mean number of bacterial infections in hospitals were same for the various colony counts from different patients.

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