A study is designed to investigate whether there

Steven Pugh

Steven Pugh

Answered question

2022-03-15

A study is designed to investigate whether there is a difference in response to various treatments in patients with Rheumatoid arthritis. The outcome is the patient’s self-reported effect of treatment. The data are shown below.
 Symptoms Worsened  No Effect  Symptoms Improved  Total  Treatment 1 22141450 Treatment 2 14152150 Treatment 3 9122950
Is there a significant difference in effect of treatment? Run the test at a 5% level of significance.

Answer & Explanation

Rowan Waller

Rowan Waller

Beginner2022-03-16Added 2 answers

For the given data, the chi-square test for association is appropriate.
Null Hypothesis:
There is no association between treatment and result.
Alternative Hypothesis:
There is an association between treatment and result.
After install the MegaStat Add-In into EXCEL, enter the given data in EXCEL sheet as shown below:
 Symptoms Worsened  No Effect  Symptoms Improved  Treatment 1 221414 Treatment 2 141521 Treatment 3 91229
Step-by-step procedure to obtain the chi-square test statistic using MegaStat:
1. Go to MegaStat > Chi-square/Crosstab > Contingency table.
2. Select the input range of the data as “Sheet1!$A$1:$D$4”.
3. In output options, Select expected values and Chi-square.
4. Click OK.
From the MegaStat output, the chi-square test statistic is 11.36 and p-value is 0.0228, which is given in the below output:
 Symptoms Worsened  No Effect  Symptoms Improved  Total  Treatment 1  Observed 22141450 Expected 15.0013.6721.3350.00 Treatment 2  Observed 14152150 Expected 15.0013.6721.3350.00 Treatment 3  Observed 9122950 Expected 15.0013.6721.3350.00 Total  Observed 454164150 Expected 45.0041.0064.00150.00
11.36 chi-square
4 df
0.228 p-value
Decision rule:
If p-valueα, then reject the null hypothesis. Otherwise, fail to reject the null hypothesis.
Conclusion:
Here, p-value (=0.0228) is less than the significance level (0.05).
Therefore, reject the null hypothesis.
There is no sufficient evidence to say that the treatment is significantly effective at 0.05 level of significance.

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