Question

An experiment is performed to determine the effects of storage length and relative humidity on the viability of seeds. Sixty-three batches of 300 seeds each are randomly divided into 21 groups of three. These 21 groups each receive a different treatment, namely the combinations of storage length (0, 60, 120, 180, 240, 300, or 360 days) and storage relativehumidity (032, or 45%). After the storage time, the seeds are planted, and the response is the percentage of seeds that sprout. The data is provided in the table below. Storage length in days Humidity 0 60120 180 240 300 360 096 | 82.1 | 78.6 | 79.8 | 82.3 | 81.7 | 85.0 82.7 79.0 80.8 79.1 75.5 80.1 879 846 81.9 80.5 78.2 79.1 81.1 82.1 81.7 32% 83.1 781 80.4 77.8 83.8 82.0 81.0 80.5 83.6 81.8 80.4 83.7 77.6 789 82.4 78.3 83.8 78.8 81.5 80.3 83.1 45% 83.1/ 66.5 52.9152.9152·2138.61252 78.9 61.4 58.9 54.3 51.9 37.9 25.8 | 81.0 | 61.2 | 59.3 | 4871 48.8140.6121 ー (use Minitab) Analyze the data to see which factor / interaction has an effect on the supporting percentage Give an interpretation of the above results. What major recommendation can you provide? a) b) c)
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Here we need to check for there will be any interaction effect between length and humidity presence or not , for this we will use MINITAB software, and the anova table is

Two-way ANOVA: y versus Humidity, length

Source DF SS MS F P
Humidity 2 41834 20917.2 2.54 0.091
length 6 51462 8577.0 1.04 0.412
Interaction 12 102725 8560.4 1.04 0.432
Error 42 345419 8224.3
Total 62 541441

S = 90.69 R-Sq = 36.20% R-Sq(adj) = 5.82%

a) here interaction effect p-value =0.432

so we can say there will be no interaction effect with the huminity and length since p-vale>0.05

b) we can say that the humidity is not related with length so that there will be no joint effect of them

c) in that case we may can follow one way ANOVA for both the humidity and length for that also the conclusion will be same.

Add a comment
Know the answer?
Add Answer to:
An experiment is performed to determine the effects of storage length and relative humidity on the...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 6.15. Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty...

    6.15. Parts manufactured by an injection molding process are subjected to a compressive strength test. Twenty samples of five parts each are collected, and the compressive strengths (in psi) are shown in Table 6E.11. (a) Establish 7 and R control charts for compressive strength using these data. Is the process in statis- tical control? (b) After establishing the control charts in part (a), 15 new subgroups were collected and the com- pressive strengths are shown in Table GE.12. Plot the...

  • Steps Open the Heart Rate Dataset in Excel Using the classification of variables from Unit 1 assignment as qualitativ...

    Steps Open the Heart Rate Dataset in Excel Using the classification of variables from Unit 1 assignment as qualitative, quantitative discrete, or quantitative continuous, match each of the 3 variables to the most appropriate graph type. (For example, qualitative data can best be displayed with a pie chart or bar graph; continuous numerical data can best be displayed using a histogram) Use the graphing functions in Excel to create an appropriate graph of the data for each variable. Remember to...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT