1 answer

Please show full details steps for better understanding. Thank you. 5. An experiment was conducted to...

Question:

Please show full details steps for better understanding. Thank you.

5. An experiment was conducted to measure and compare the effectiveness of various feed supplements on the growth of chickens

ANOVA table ANOVA Table Mean Square F Value Pr>F BFB Model: weight-feed HO: The means for all levels are equal Source Sum of

5. An experiment was conducted to measure and compare the effectiveness of various feed supplements on the growth of chickens. Use the software output to answer the following questions: Chick Weights 400 casei - horsebean Anseed meatmeal soybean sunflower 350 SUD Weight OCZ COZ 150 cybean berben Feed Type (a) [6 points] Use the boxplots to assess the conditions for testing if the mean weights differ across feed type.
ANOVA table ANOVA Table Mean Square F Value Pr>F BFB Model: weight-feed HO: The means for all levels are equal Source Sum of DF Squares feed 5 2.3113e+05 Residual 65 1.9556e+05 feed is significant at 5% significance level. 46226 15.365 5.9364e-10 2.917e+07 3008.6 (b) [10 points] Use the software output table to write the out the 4-steps of hypothesis test for testing if the mean weights differ across feed type. (c) (4 points) Interpret the meaning of this p-value in context of the problem.

Answers

There are 6 feed suppliments on the growth of chickens

The weights are considered for all 6 feeds that is treatments.

a) Using the box plot,

1. The samples taken for each suppliment are simple random samples.

2. The samples taken for each suppliment are independent on each other that is each feed type is independent on each other.

3. The populations from which the samples of feed are selected are normal.

4. The variances of the populations from which the samples are selected are equal.

b) 4 Steps of hypothesis

1.

The null and alternative hypothesis

H0: The means for all levels are equal

H1: Atleat one mean differ from the others.

2. The test statistics

Uisng the software output, the F test statistics is

F value = 15.365

3. P-value

Uisng the output,

P-value = 5.9364e-10 = 0.00000000059364

4. Decision and conclusion

Alpha is not given so it should be 0.05

If P-value > alpha then fail to reject the null hypothesis otherwise reject the null hypothesis

P-value (0.00000000059364) is less than alpha 0.05 so reject the null hypothesis.

Conclusion: Reject the null hypothesis, there is sufficient evidence to support that claim that at least one mean differs across feed type.

c) P-value

P-value is a probability of getting a value of test statistics as more as extreme than the value of test statistics computed from the data, under the assumption that the null hypothesis it true.

P-value is the smallest level of significance which lead to reject the null hypothesis.

That is if the p-value is too small then null would be rejected and if not then it would be fail to rejected.

In this case p-value is too small so it leads to reject the null hypothesis.

.

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