A statistician is conducting a chi-square goodness-of-fit test and is limited by the cost, per individual, to conduct the study. The statistician selects a sample of size 39, which is the smallest sample possible that will meet the condition for large expected counts. Which of the following could not be the null hypothesis for the study? A) H0: p1=0.20, p2=0.20, p3=0.20, p4=0.20, p5=0.20 B) H0: p1=0.15, p2=0.35, p3=0.22, p4=0.15, p5=0.13 C) H0: p1=0.24, p2=0.23, p3=0.21, p4=0.18, p5=0.14 D) H0: p1=0.34, p2=0.21, p3=0.14, p4=0.15, p5=0.16 E) H0: p1=0.43, p2=0.23, p3=0.17, p4=0.09, p5=0.08
A statistician is conducting a chi-square goodness-of-fit test and is limited by the cost, per individual, to conduct the study. The statistician selects a sample of size 39, which is the smallest sample possible that will meet the condition for large expected counts. Which of the following could not be the null hypothesis for the study?
A) H0: p1=0.20, p2=0.20, p3=0.20, p4=0.20, p5=0.20
B) H0: p1=0.15, p2=0.35, p3=0.22, p4=0.15, p5=0.13
C) H0: p1=0.24, p2=0.23, p3=0.21, p4=0.18, p5=0.14
D) H0: p1=0.34, p2=0.21, p3=0.14, p4=0.15, p5=0.16
E) H0: p1=0.43, p2=0.23, p3=0.17, p4=0.09, p5=0.08
Answered
Answer: E) H0: p1=0.43, p2=0.23, p3=0.17, p4=0.09, p5=0.08
Explanation: For a sample size of 39, each expected count should be at least 5. With these proportions, some of the expected counts would be less than 5, making this hypothesis not feasible.
BNO Team