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In order to evaluate the effectiveness of a new type of plant food that was developed...

In order to evaluate the effectiveness of a new type of plant food that was developed for tomatoes, a study was conducted in which a random sample of n = 76 plants received a certain amount of this new type of plant food each week for 14 weeks. The variable of interest is the number of tomatoes produced by each plant in the sample. The table below reports the descriptive statistics for this study: Descriptive statistics for the tomato food study Variable n sample mean sample standard deviation standard error number of tomatoes 76 33.40 8.60 0.986 Assuming the population is normally distributed, the investigators would like to construct a 98% confidence interval for the average number of tomatoes that all plants of this variety can produce when fed this supplement like this. a) The margin of error is: 2.426 Correct: Your answer is correct. (3 decimals) b) The corresponding 98% confidence interval for the true population mean is: Lower Limit: Incorrect: Your answer is incorrect. (3 decimals) to Upper Limit: Incorrect: Your answer is incorrect. (3 decimals) c) What would we conclude at α = 0.02 for the hypothesis test H0: μ = 37.845 vs. Ha: μ ≠ 37.845? We do not have enough evidence to conclude the true mean is 37.845. We have sufficient evidence to conclude that the true mean is different from 37.845. We have insufficient evidence to conclude the true mean is different from 37.845. We have enough evidence to conclude that the true mean is 37.845. We do not have enough evidence to conclude the true mean is 33.40.  

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Answer #1

Given X bar = 33.40, Sd = 8.60, SE = 0.986 and n = 76

C = 98%, Alpha = 1-c = 0.02 and df = n-1 = 75

tc = 2.3771 (Use t table)

a) margin of Error ME = tc*SE = 2.3771*0.986 = 2.3438

b)

CI
tc 2.377102 T.INV.2T(D1,D2)
Upper 35.745 X bar + tc*(S/SQRT(n))
Lower 31.055 X bar - tc*(S/SQRT(n))

98% CI = (31.055, 35.745)

c)

Hypothesis

H0: μ = 37.845

Ha: μ ≠ 37.845

Test :
t -4.506 (X bar-μ )/(S/SQRT(n))
P value 2.389E-05 T.DIST.2T(-ts,df)

P value < 0.02,Rejrct H0

We have sufficient evidence to conclude that the true mean is different from 37.845

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