Bird eggs A researcher wants to see if birds that build larger nests lay larger eggs. She selects two random samples of nests: one of the small nests and the other of large nests. Then she weighs one egg (chosen at random if there is more than one egg) from each nest. A 95% confidence interval for the difference (Large − Small) between the mean mass (in

grams) of eggs in small and large nests is 1.6±2.0 role="math" localid="1654714720708" 3051526=0.200=20.0%1.6±2.0

a. Does the interval provide convincing evidence of a difference in the true mean egg mass of birds with small nests and birds with large nests? Explain your answer.

b. Does the interval provide convincing evidence that the true mean egg mass of birds with small nests and birds with large nests is the same? Explain your answer.

Short Answer

Expert verified

Part a) No

Part b) No

Step by step solution

01

Part a) Step 1: Explanation

The difference in mean egg mass between birds with small nests and birds with large nests has a 95%confidence interval of:

role="math" localid="1654715011426" 1.6±2.0=(1.6-2.0,1.6+2.0)=(-0.4,3.6)

Then we notice that the confidence interval contains zero, implying that the difference in mean reaction time is likely to be zero, and thus that no difference exists.

This means that there is no convincing evidence of a difference in true mean egg mass between birds with small and large nests.

02

Part b) Step 1: Explanation

The difference in mean egg mass between birds with small nests and birds with large nests has a 95%confidence interval of:

1.6±2.0=(1.6-2.0,1.6+2.0)=(-0.4,3.6)

Then we notice that the confidence interval contains zero, implying that the difference in mean reaction time is likely to be zero, and thus that no difference exists.

There is no evidence that the true mean egg mass of birds with small nests and birds with large nests is the same because the difference of zero is one of the plausible values, but the confidence interval for the difference contains many other plausible values.

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Most popular questions from this chapter

Based on the P-value in Exercise 71, which of the following must be true?

a. A 90%confidence interval for μM−μF3051526=0.200=20.0%μM-μFwill contain 0

b. A 95%confidence interval for μM−μF3051526=0.200=20.0%μM-μFwill contain 0

c. A 99%confidence interval for μM−μF3051526=0.200=20.0%μM-μFwill contain 0

d. A 99.9% confidence interval for μM−μF3051526=0.200=20.0%μM-μF will contain 0

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c. Do the data provide convincing evidence at the α=0.01,3051526=0.200=20%significance level that subliminal messages help students like the ones in this study learn math, on average?

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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.

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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)

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e. No conclusion can be drawn because the other coordinates are unknown.

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