B) Normal probabilty plots indicate that the sample data come from normal popula

Question

B) Normal probabilty plots indicate that the sample data come from normal populations. Are the requirements to use the one-way ANOVA procedure satisfied?

1) No, because the largest sample standard deviation is more than twice the smallest sample standard deviation.

2) No, because there are K=3 simple random samples, one from each of K populations. The K samples are independent of each other, and the populations are normally distrubted and have the same variance.

3) Yes, Because there are K=3 simple random samples, one from each of K populations. The K Samples are independent of each other, and the populations are normally distrubted and have different vacanies.

4) Yes, Because there are K=3 simple random samples, one from each of K populations. The K samples are independent of each other, and the populations are normally distrubted and have the same variance.

C) Are the mean rates of return different at a=0.05 level of significane?

Use the technology to find the F-test statistic for this data set

Fn= ___

An engineer wants to know if the mean strengths of three different concrete mix designs differ significantly. He randomly selects 9 cylinders that measure 6 inches in diameter and 12 inches in height in which mixture A is poured, 9 cylinders of mixture B, and 9 cylinders of mixture C. After 28 days, he measures the strength (in pounds per square inch) of the cylinders. The results are presented in the accompanying table. Click the icon to view the data table. (a) State the null and alternative hypotheses. Choose the correct answer below O B. Ho: A-Ps and H1 : at least one of the means is different O C. Ho: at least one of the means is different and H1 : A-8-c O D. Ho: HA B and H1 the means are different

Explanation / Answer

question 1

Here Option B is correct. as null hypothesis conveys that all mean are same and alternative hypothesis entails that one of the means is different.

Question 2

Option 4 is correct here as homogenity of variance is a requirement here.

Question 3

From using Excel. the anova analysis is given below.

F = 3.7424 here  

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Mix A 9 35620 3957.778 11219.44 Mix B 9 38480 4275.556 151502.8 Mix C 9 36230 4025.556 39377.78 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 504229.6 2 252114.8 3.742427 0.038487 3.402826 Within Groups 1616800 24 67366.67 Total 2121030 26

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

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