Question

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select$10$Variety A and$10$Variety B tomato plants. Then the researchers divide in half each of$10$small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The$10$differences (Variety A − Variety B) in yield are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. The mean difference is ${\stackrel{-}{x}}_{A-B}$$=0.34$$\frac{305}{1526}=0.200=20\%{\stackrel{-}{x}}_{A-B}=0.34$and the standard deviation of the differences is s A-B$=0.83$$\frac{305}{1526}=0.200=20\%=s_{A-B}=0.83$.Let ${\mu }_{A-B}=\frac{305}{1526}=0.200=20\%$μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

A 95% confidence interval for${\mu }_{A-B}\frac{305}{1526}=0.200=20\%{\mu }_{A-B}$is given by

a. $0.34\pm 1.96(0.83)\frac{305}{1526}=0.200=20\%0.34\pm 1.96(0.83)$

b.$0.34\pm 1.96(0.8310)\frac{305}{1526}=0.200=20\%0.34\pm 1.96(0.8310)$

c. $0.34\pm 1.812(0.8310)\frac{305}{1526}=0.200=20\%0.34\pm 1.812(0.8310)$

d. $0.34\pm 2.262(0.83)\frac{305}{1526}=0.200=20\%0.34\pm 2.262(0.83)$

e.$0.34\pm 2.262(0.8310)\frac{305}{1526}=0.200=20\%0.34\pm 2.262(0.8310)$

Short answer

The true mean difference is$0.34\pm 2.262(0.8310)$and the correct option is (e)

Step-by-step solution

Given Information

We are given the true difference and we have to find out which value will be satisfied from the given options.

Explanation

According to the question,

sample size is$10$, mean difference is$0.34$, standard deviation difference is$0.83$and confidence level is$0.95$.

To find the degree of freedom df$=10-1=9$and $t_{\frac{\alpha }{2}}=2.262$and put these values to find the margin of error, which is $E=t_{\alpha /2}\times \frac{s}{\sqrt{n}}=2.262\times \frac{0.83}{\sqrt{10}}$

and as the boundaries of confidence is$\stackrel{-}{x}\pm E$.

Put all the values in the expression,

We get, $0.34\pm 2.62\times \frac{0.83}{\sqrt{10}}$.

Hence, option (e) is correct.