Two samples or paired data? In each of the following settings, decide whether you should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference. Explain your choice.

a. To test the wear characteristics of two tire brands, A and B, each of 50cars of the same make and model is randomly assigned Brand A tires or Brand B tires.

b. To test the effect of background music on productivity, factory workers are observed. For one month, each subject works without music. For another month, the subject works while listening to music on an MP3 player. The month in which each subject listens to music is determined by a coin toss.

c. How do young adults look back on adolescent romance? Investigators interviewed a random sample of 40couples in their mid-twenties. The female and male partners were interviewed separately. Each was asked about his or her current relationship and also about a romantic relationship that lasted at least 2months when they were aged 15or 16. One response variable was a measure on a numerical scale of how much the attractiveness of the adolescent partner mattered. You want to find out how much men and women differ on this measure.

Short Answer

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Part(a) We should use two-sample t procedures to perform inference about a difference in means.

Part(b) We should use paired t procedures to perform inference about a mean difference.

Part(c) We should use paired t procedures to perform inference about a mean difference.

Step by step solution

01

Part(a) Step 1 : Given information

We need to decide whether to use two-sample t procedures or paired t procedures to check inference about a mean difference.

02

Part(a) Step 2 : Simplify

We must use paired t techniques if the two samples contain the same individuals or if the subjects in one sample are related to the subjects in the other sample.
We must use two sample t techniques if the subjects in the two samples are completely unrelated.
Automobiles are randomly assigned to either brand A or brand B in this scenario, resulting in a first sample of brand A cars and a second sample of brand B cars.
Because the cars were randomly assigned to one of the samples, the cars in the two samples will be completely unrelated, making the two sample t methods suitable.


03

Part(b) Step 1 : Given information

We need to decide whether we should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference.

04

Part(b) Step 2 : Simplify

If the two samples contain the same individuals or if the subjects in one sample are connected to the subjects in the other sample, we must employ paired t methods.
If the subjects in the two samples are fully unrelated, we must employ two sample t methods.
For one month, each topic worked with music and for one month, each subject worked without music.
The data for all participants who worked with music for one month is the first sample, while the data for all subjects who worked without music for one month is the second sample.
We should utilise paired t methods because the two samples are the same subjects.

05

Part(c) Step 1 : Given information

We need to decide whether we should use two-sample t procedures to perform inference about a difference in means or paired t procedures to perform inference about a mean difference.

06

Part(c) Step 2 : Simplify

If the two samples contain the same individuals or if the subjects in one sample are connected to the subjects in the other sample, we must employ paired t methods.
If the subjects in the two samples are fully unrelated, we must employ two sample t methods.
We questioned the male and female partners separately in each of the 40couples.
The male partners are in the first sample, while the female partners are in the second.
Because all of the participants in the first sample are male spouses of a subject in the second sample, we should utilise the paired t methods to compare the two samples.

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Most popular questions from this chapter

On your mark In track, sprinters typically use starting blocks because they think it will help them run a faster race. To test this belief, an experiment was designed where each sprinter on a track team ran a 50-meter dash two times, once using starting blocks and once with a standing start. The order of the two different types of starts was determined at random for each sprinter. The times (in seconds) for 8 different sprinters are shown in the table.

a. Make a dotplot of the difference (Standing - Blocks) in 50-meter run time for each sprinter. What does the graph suggest about whether starting blocks are helpful?

b. Calculate the mean difference and the standard deviation of the differences. Explain why the mean difference gives some evidence that starting blocks are helpful.

c. Do the data provide convincing evidence that sprinters like these run a faster race when using starting blocks, on average?

d. Construct and interpret a 90%confidence interval for the true mean difference. Explain how the confidence interval gives more information than the test in part (b).

TicksLyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population, in turn, rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. To see if mice are more likely to breed when there are more acorns, the researchers added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54of the 72mice trapped in the first area were in breeding condition, versus 10of the 17mice trapped in the second area.

a. State appropriate hypotheses for performing a significance test. Be sure to define the parameters of interest.

b. Check if the conditions for performing the test are met.

Music and memory Refer to Exercise 87.

a. Construct and interpret a 99%confidence interval for the true mean difference. If you already defined the parameter and checked conditions in Exercise87, you don’t need to do them again here.

b. Explain how the confidence interval provides more information than the test in Exercise .

A random sample of size n will be selected from a population, and the proportion p^3051526=0.200=20.0%p^ of those in the sample who have a Facebook page will be calculated. How would the margin of error for a 95% confidence interval be affected if the sample size were increased from 50to200 and the sample proportion of people who have a Facebook page is unchanged?

a. It remains the same.

b. It is multiplied by 2.

c. It is multiplied by 4.

d. It is divided by 2.

e. It is divided by 4.

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-=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.LetμA-B=3051526=0.200=20%μA−B = the true mean difference (Variety A − Variety B) in yield for tomato plants of these two varieties.

The P-value for a test of H0: μA−B=03051526=0.200=20%versus Ha: μA−B≠0 is 0.227. Which of the following is the

correct interpretation of this P-value?

a. The probability that μA−B is0.227.

b. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34is0.227.

c. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference of0.34or greater is0.227.

d. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is0, the probability of getting a sample mean difference greater than or equal to0.34or less than or equal to −0.34is0.227.

e. Given that the true mean difference (Variety A – Variety B) in yield for these two varieties of tomato plants is not 0, the probability of getting a sample mean difference greater than or equal to 0.34or less than or equal to −0.34is0.227.

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