QUESTION 4 A soil scientist has just developed a new type of fertilizer and she

Question

QUESTION 4 A soil scientist has just developed a new type of fertilizer and she wants to determine whether it helps carrots grow lar added to half the pots Alilthe pots are placed in a temperature-controlled greenhouse where they receive adequate light and equal amounts of water After two months of growth, the scientist harvests the carrots ger. She sets up several pots of soil and plants one carrot seed in each pot Fertilizer is them fin kilograms) Below is a data table showing the weight of the carrots at the end of the growing period from the two treatment groups. Here is a When analyzing this dataset with a t-test, the states that the fertilized carrots are larger in size To ana?yze this data set, should the scientist use a hypothesis states that the average size of the carrots from each treatment are the same, whereas the hypothesis tailed t-test? After performing a t-test assuming equal variances using the data analysis ads-in for MS Excel, what is the calculated t-value for this data set? [ . Round your answer to four decimal places. Your answer should be a positive value Report the appropiate critical t value, based on your decision of a one- or two-tailed test, calculated by MS Excel using the Data Analysis add-in. Round your answer to four decimal places What is the appropriate p value for the t-test? . Report your answer in exponential notation . Report your answer to 4 decimal places after converting to exponential notation e g o 1.1234E-01 for 0 112341 o 3.1234E-04 for 0.000312341 Would you reject or fail to reject the null hypothesis? Chick Save and Submit to save and submit. Click Save All Answers to save all answers.

Explanation / Answer

When analyzing the data set with t-test, null hypothesis states that average size of the carrots from each treatment are the same whereas alternate hypothesis states that fertilized carrots are larger in size.

Scientist should use a one tailed t test

Appropriate p value for the t test is 0.05

It is a independent/ unpaired t test that compare mean of two group. t test tells how significant the difference are. To calculate t test first we have to calculate mean of each group than minus each value from their respective mean and square that value and take sum of square.

Here

Control group mean M1 = 0.36

Degree of freedom n1 = N-1 = 30-1 = 29

Sample square (£X-M1) = 1.09

Treatment group M2 = 0.5

Sample square ( £Y-M2) = 1.09

Degree of freedom n2 = 29

Calculate standard deviation (SD)

? £( X-M1)^2 + £(Y-M2)^2/N1+N2-2

Here, ?0.27+1.09/30+30-2

= 0.1531

Standard error (SE) = SD ?1/N1+1/N2

0.1531?1/30+1/30

0.1531 ?0.06

0.1531* 0.2449

0.0375

t = M1-M2/SE

Calculated t value = -3.6211

p value = .000309

Here, the result is significant at p<0.05. we reject the null hypothesis. There is significant difference between the mean of the two group.

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