Suppose that Charlotte is a medical researcher at a health and wellness company
ID: 3326856 • Letter: S
Question
Suppose that Charlotte is a medical researcher at a health and wellness company who wants to test the effectiveness of a new sleeping pill by comparing test and control groups. She plans to to run a hypothesis test at a significance level of =0.01 and randomly selects 286 patients with insomnia from patient databases at doctors' offices in the area. She randomly assigns 158 patients to the test group and 128 patients to the control group. The test group receives the new test treatment while the control group receives a standard OTC (over the counter) treatment.
Charlotte follows up with the patients one month after the treatment administration to evaluate their sleep quality. She records the patients' answers to a survey question regarding the number of nights of poor sleep quality per week since the start of the treatment. She categorizes the results as insomnia classifications of mild, moderate, significant, and severe. Charlotte proceeds to run a chi-square test of homogeneity to determine if the proportion of patients in each insomnia category is the same between the new treatment and standard treatment groups. The sample results are summarized in the contingency table.
2=4.993
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Compute the p-value for the chi-square statistic, 2, for Charlotte's hypothesis test using software. You may find this list of software manuals helpful. Provide your answer with precision to three decimal places. Avoid rounding until the final step.
p-value =
Select the accurate statement regarding Charlotte's hypothesis test decision and conclusion.
Charlotte should reject the null hypothesis because there is sufficient evidence (p-value > 0.01) to reject the null. Thus, Connor should conclude that at least one of the proportions of patients for one of the insomnia ratings is different between the test treatment group and standard treatment group.
Charlotte should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.01) to reject the null. Thus, Charlotte should conclude that the insomnia classification proportions are uniform between the test treatment group and standard OTC treatment group.
Charlotte should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.01) to reject the null. Thus, Charlotte cannot conclude whether or not insomnia classification proportions are uniform or different between the test treatment group and standard OTC treatment group.
Charlotte should reject the null hypothesis because there is sufficient evidence (p-value < 0.01) to reject the null. Thus, Charlotte should conclude that at least one of the proportions of patients for an insomnia classification is different between the test treatment group and standard treatment group.
Charlotte should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.01) to reject the null. Thus, Charlotte should conclude that at least one of the proportions is significantly different between the treatment groups.
Mild(< 2 nights
poor sleep
per week) Moderate
(2–3 nights
poor sleep
per week) Significant
(4–5 nights
poor sleep
per week) Severe
(> 5 nights
poor sleep
per week) All New
treatment Observed count 81 58 11 8 158 Expected count 85.63 50.27 14.36 7.73 Chi-square 0.25028 1.18774 0.78769 0.00913 Standard
treatment Observed count 74 33 15 6 128 Expected count 69.37 40.73 11.64 6.27 Chi-square 0.30894 1.46611 0.97230 0.01127 All 155 91 26 14 286
Explanation / Answer
Solution:
Using excel
=CHISQ.DIST.RT(4.993,3)
= 0.1723
Since p-value is greater than 0.01, we fail to reject the null hypothesis.
Charlotte should fail to reject the null hypothesis because there is insufficient evidence (p-value > 0.01) to reject the null. Thus, Charlotte should conclude that the insomnia classification proportions are uniform between the test treatment group and standard OTC treatment group.