The file \"SWISS.xlsx\" (attached here) contains data on serum creatinine levels

Question

The file "SWISS.xlsx" (attached here) contains data on serum creatinine levels according to NAPAP group. The urine N-acetyl P-aminophenyl (NAPAP) level was used as a marker of phenacetin-containing analgesic intake. Participants in the study were divided into three groups: group 1 = high NAPAP, group 2 = low NAPAP, group 3 = control group. Use this dataset to conduct a test of the null hypothesis H0: is mean serum creatinine eqaul across the high NAPAP, low NAPAP and control groups?

What is the F test statistic for testing the hypothesis of interest?

What is the associated P-Value?

What is the conclusion of this test?

Group 1 Group 2 Group 3 0.97 0.7 1.48 0.88 0.84 0.78 0.96 0.47 0.79 0.74 0.81 0.8 0.85 0.75 0.84 1.06 0.76 0.94 0.65 0.91 0.86 0.65 0.87 1.5 0.94 0.99 0.8 0.86 0.85 0.88 0.78 0.66 0.78 0.84 0.85 1.2 0.75 0.77 1 0.82 0.74 0.72 0.76 0.89 0.76 2.1 0.9 0.66 1.38 0.61 0.71 1.8 0.75 0.75 0.77 1.04 0.72 0.76 0.8 1.19 1.2 0.66 0.9 0.82 0.75 9.99 0.99 0.72 0.6 0.71 0.96 0.66 0.92 0.73 0.65 0.75 0.82 0.65 0.85 0.68 0.81 3.25 0.74 0.56 1.25 1.41 0.84 1.07 0.77 0.72 0.85 0.71 0.85 9.99 0.98 1.01 1.48 0.58 0.71 0.84 0.9 0.83 0.95 0.87 0.75 0.68 9.99 0.83 0.75 1.03 0.82 0.95 0.82 0.81 0.75 0.82 0.87 1.9 0.86 0.71 0.85 1.29 0.77 9.99 1.45 0.67 1.17 1.05 1.22 0.75 1 0.73 0.87 0.86 1.24 1 0.67 0.82 0.75 1.55 0.62 1.25 0.75 0.75 1.15 0.8 0.93 1.45 0.72 0.9 1.35 0.9 0.93 0.95 0.88 9.99 1.05 0.89 0.67 1 0.72 0.87 0.8 0.96 0.68 2.05 0.85 0.7 1.25 0.95 0.87 1.75 0.95 0.92 0.95 1.55 0.94 1.45 0.65 0.9 1.15 0.95 0.77 1.2 1.1 0.9 1.02 1.05 0.71 1.15 1.15 0.93 0.9 0.95 0.72 1.15 1.25 1.15 1.4 1.35 0.76 0.7 0.85 1.5 0.7 0.8 0.99 0.75 0.85 0.85 1.36 0.45 1.23 0.81 1.1 1.5 0.98 0.9 1.2 0.7 1 0.69 0.7 0.75 0.71 0.79 0.85 1.2 0.78 0.75 0.94 0.85 1.03 0.84 0.9 1 0.94 0.94 0.95 0.93 1.05 0.8 1.28 3.83 0.83 0.9 1.15 0.96 1.09 1.2 0.85 0.84 0.96 0.98 1.25 1.04 0.69 0.83 0.75 0.73 0.94 0.71 0.85 0.55 1.56 1.23 0.85 0.92 0.91 0.76 1.19 1.57 0.95 1.26 1.48 0.9 0.87 1.63 0.72 1.24 1.28 0.8 1.39 0.99 0.95 0.85 0.94 0.92 1.16 0.89 1.8 1.5 0.67 0.8 0.96 1.25 0.7 0.69 0.74 1.2

Explanation / Answer

Null Hypothesis H0: mew1=mew2=mew3 i.e there is no significance difference between the mean serum creatinine the high NAPAP, low NAPAP and control groups( all are equal)

Alternative Hypothesis H1:atleast two are not equal(two-tailed test)

               Descriptives
Treatment
  
N   Mean   Std. Deviation   Std. Error
Group1   100   1.2531   1.33896 .13390   
Group2   100   1.0055   .93726 .09373
Group3   100   1.0734   1.29984 .12998   
Total 300   1.1107   1.20617 .06964   

Test of Homogeneity of Variances
Treatment
Levene Statistic   df1   df2   Sig.
1.241   2   297   .291

       ANOVA One-way classification table
Treatment
Sum of Squares   df   Mean Square   F Sig.
Between Groups   3.274 2 1.637 1.126   .326
Within Groups   431.723 297 1.454      
Total   434.997 299          

F test statistic for testing the hypothesis of interest = 1.126

the associated P-Value = The p-value is 0.325708. The result is not significant at p(0.325708) > 0.05

i.e H0 is accepted i.e we may conclude that there is no significance difference between the mean serum creatinine the high NAPAP, low NAPAP and control groups( all are equal)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.