Question

The following data represent quantities of tea leaf pluckings (tender shoots from tea plants) from sixteen different plots of tea bushes intended for experimental use in Ceylon, a type of tea from Sri Lanka. The tea bushes are randomly divided into four different treatment groups. Experimenters wish to determine if the mean number of pluckings differs among the four treatments. Test this using 5% significance, assuming that these samples are drawn from normal populations with equal variances. Treatment 1 Treatment 2 Treatment 3 Treatment Pluckings Pluckings Pluckings Pluckings 88 102 91 88 94 110 109 118 109 110 115 94 88 102 91 961. Give the null and alternative2. What is the value of the F-test statistic? Add the F-test statistic to the graph and shade the right tail. Use technology to determine the P-value for this hypothesis test. 3. What does the P-value tell you about the null and alternative hypotheses? 4. State a conclusion in the context of this problem.

Answer 1

Answer:

Explanation:

1)$H_0$ : All group means equal or $\mu _1=\mu _2=\mu _3=\mu _4$

$H_1:\mu 1=\mu 2=\mu 3\neq \mu 4$

At least one of the treatment group means are different

ANOVA TABLE      

Source of Variation    SS         df      MS            F          P-value        F crit

Between Groups       213.5      3    71.16667   0.65 0.5975     3.490295

Within Groups          1312.5      12   109.375  

MSB = SSB / DFB = 71.16667

MSE = SSE / DFE = 109.375

F = MSB / MSE = 0.650667

3) P-value: 0.597576

The test statistic is not significant and failed to reject the null hypothesis.

4) The test statistic is not significant. So, there is no evidence to conclude that there is a difference between groups.