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Better barley Does drying barley seeds in a kiln increase the yield of barley? A famous experiment by William S. Gosset (who discovered the t distributions) investigated this question. Eleven pairs of adjacent plots were marked out in a large field. For each pair, regular barley seeds were planted in one plot and kiln-dried seeds were planted in the other. A coin flip was used to determine which plot in each pair got the regular barley seed and which got the kiln-dried seed. The following table displays the data on barley yield (pound per acre) for each plot.

Do these data provide convincing evidence at the α=0.05level that drying barley seeds in a kiln increases the yield of barley, on average?

Short Answer

Expert verified

There is no convincing evidence that drying barley seeds in a kiln increase the yield of barley, on average.

Step by step solution

01

Given information

We were told that on a big field, eleven pairs of neighboring plots were marked out, with ordinary barley seeds planted in one plot and kiln-dried seeds placed in the other.

We need to find out that do these data provide convincing evidence at the α=0.05level that drying barley seeds in a kiln increases the yield of barley, on average

02

Explanation

Given:

n=Samplesize=11α=Significancelevel=0.05

Let us determine the difference between regular barley seeds and kiln-dried seeds

Now we will determine the mean of values of the difference:

x¯=-106+20-101+33-72-62+36-38+70-127-2411=-37111≈-33.7273

Now we will determine the standard deviation

s=∑n=110Difference-x¯2n-1

s=106+33.72732+20+72732+-101+33.72732+33+33.72732+-72+33.72732+-62+33.72732+36+33.72732+-38+33.72732+70+33.72732+-127+3.72732+-24+33.7273211-1≈66.1711

Now we will carry out a hypothesis test for the population mean difference.

Here we have:

Populationmeandifference=μdH0=NullhyphothesisHa=AlternativehyphothesisH0:μd=0Ha:μd≠0Nowwewilldeterminethevalueofteststatistics:t=x¯-μdsn

t=-33.7273-066.171111≈-1.690

The P-value is the probability of obtaining the value of test statistics.

Degree of freedom =11-1=10

The test is a two-tailed test so we double the boundaries of the P-value.

0.05<P<0.10nowatt=-1.690andDegreeoffreedom=10,Pvalue=0.06095

We have to reject the null hypothesis if the probability value is less than the hypothesis value.

P>0.05⇒failtorejectH0

This demonstrates that there is no convincing evidence that kiln-drying barley seeds increases barley yields on average.

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Most popular questions from this chapter

An agricultural station is testing the yields for six different varieties of seed corn. The station has four large fields available, located in four distinctly different parts of the county. The agricultural researchers consider the climatic and soil conditions in the four parts of the county as being quite different, but are reasonably confident that the conditions within each field are fairly similar throughout. The researchers divide each field into six sections and then randomly assign one variety of corn seed to each section in that field. This procedure is done for each field. At the end of the growing season, the corn will be harvested, and the yield (measured in tons per acre) will be compared. Which one of the following statements about the design is correct?

a. This is an observational study because the researchers are watching the corn grow.

b. This a randomized block design with fields as blocks and seed types as treatments.

c. This is a randomized block design with seed types as blocks and fields as treatments.

d. This is a completely randomized design because the six seed types were randomly assigned to the four fields.

e. This is a completely randomized design with24³Ù°ù±ð²¹³Ù³¾±ð²Ô³Ù²õ—6 seed types and 4 fields.

Mrs. Woods and Mrs. Bryan are avid vegetable gardeners. They use different fertilizers, and each claims that hers is the best fertilizer to use when growing tomatoes. Both agree to do a study using the weight of their tomatoes as the response variable. Each planted the same varieties of tomatoes on the same day and fertilized the plants on the same schedule throughout the growing season. At harvest time, each randomly selects 15 tomatoes from her garden and weighs them. After performing a two-sample t test on the difference in mean weights of tomatoes, they gett=5.24 andP=0.0008. Can the gardener with the larger mean claim that her fertilizer caused her tomatoes to be heavier?

a. Yes, because a different fertilizer was used on each garden.

b. Yes, because random samples were taken from each garden.

c. Yes, because the P-value is so small.

d. No, because the condition of the soil in the two gardens is a potential confounding variable.

e. No, because15<30

Researchers suspect that Variety A tomato plants have a different average yield than Variety B tomato plants. To find out, researchers randomly select10Variety A and10Variety B tomato plants. Then the researchers divide in half each of10small 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. The10differences (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 x-A-B=0.343051526=0.200=20%x-A-B=0.34and the standard deviation of the differences is s A-B=0.833051526=0.200=20%=sA-B=0.83.Let μA-B=3051526=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 forμA-B3051526=0.200=20%μA-Bis given by

a. 0.34±1.96(0.83)3051526=0.200=20%0.34±1.96(0.83)

b.0.34±1.96(0.8310)3051526=0.200=20%0.34±1.96(0.8310)

c. 0.34±1.812(0.8310)3051526=0.200=20%0.34±1.812(0.8310)

d. 0.34±2.262(0.83)3051526=0.200=20%0.34±2.262(0.83)

e.0.34±2.262(0.8310)3051526=0.200=20%0.34±2.262(0.8310)

An SRS of size 100is taken from Population A with proportion 0.8of successes. An independent SRS of size 400is taken from Population B with proportion 0.5of successes. The sampling distribution of the difference (A − B) in sample proportions has what mean and standard deviation?

a. mean=0.3; standard deviation =1.3

b. mean=0.3; standard deviation =0.40

c. mean=0.3; standard deviation =0.047

d. mean=0.3; standard deviation =0.0022

e. mean=0.3; standard deviation =0.0002

Music and memory Does listening to music while studying help or hinder students’ learning? Two statistics students designed an experiment to find out. They selected a random sample of 30students from their medium-sized high school to participate. Each subject was given 10minutes to memorize two different lists of 20words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference (Silence − Music) in the number of words recalled was recorded for each subject. The mean difference was 1.57and the standard deviation of the differences was 2.70.

a. If the result of this study is statistically significant, can you conclude that the difference in the ability to memorize words was caused by whether students were performing the task in silence or with music playing? Why or why not?

b. Do the data provide convincing evidence at theα=0.01 significance level that the number of words recalled in silence or when listening to music differs, on average, for students at this school?

c. Based on your conclusion in part (a), which type of error—a Type I error or a Type II error—could you have made? Explain your answer.

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