An advertiser is experimenting with a new color scheme and conducts a study to test its effectiveness. In which situation could they use the two-sample t-test for comparing two population means?

They randomly expose consumers to one website when they land on their page: the old one with the original color scheme or the new one with the updated color scheme. Then they measure to see how much people buy.

They track each customer's spending habits in the old platform and then they change the color scheme to see if spending habits go up or down for each consumer.

They show consumers both options: the original color scheme or the updated color scheme. They let consumers decide which one they like better and then select the color scheme that most customers prefer.

Question 2 of 5

Question 2

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College Students and Depression: A public health official is studying differences in depression among students at two different universities. They collect a random sample of students independently from each of the two universities and administer a well known depression inventory. A score of 5 or above indicates some depression. A score above 15 indicates that active treatment is necessary.

Sample Statistics

	Size(n)	Mean(x)(x)	SD(s)
Sample 1	50	9.2	.85
Sample 2	45	8.7	1.2

The official conducts a two-sample t-test to determine whether these data provide significant evidence that students at University 1 are more depressed than students at University 2. The test statistic is t = 2.64 with a P-value 0.005. Which of the following is an appropriate conclusion?

The samples provide significant evidence that students at University 1 are more depressed than students at University 2.

The samples do not provide statistically significant evidence.

We can not use the t-test in this case because the variables (depression scores) are likely skewed to the right at each university.

Question 3 of 5

Question 3

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In a study of the impact of eating fruit on birth weight, researchers analyze birth weights (in grams) for babies born to 189 women who gave birth in 2018 at a hospital in California. Suppose in the group, 74 of the women were categorized as "fruit eaters" and 115 as "non-fruit eaters." The difference in the two sample mean birth weights (fruit eaters minus non-fruit eaters) is 281.7 grams and the 95% confidence interval is (76.5, 486.9)

Which gives the best interpretation of what we can conclude about the impact of eating fruit on birth weight?

We are 95% confident that on average, eating fruit causes higher birth weights of between 76.5 grams to 486.9 grams.

There is a 95% chance that if a woman eats fruit during pregnancy her baby will weigh between 76.5 grams to 486.9 grams more than if she did not eat fruit.

Eating fruit is associated with higher birth weights. When fruit eaters are compared to non-fruit eaters, we are 95% confident that the mean weight of babies of fruit eaters is between 76.5 grams to 486.9 grams more than the mean weight of babies of non-fruit eaters.

This study does not suggest that there is a difference in mean birth weights when we compare fruit eaters to non-fruit eaters.

Question 4 of 5

Question 4

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Do dance coaches score students that they have previously taught higher at competitions due to bias? Two samples were randomly selected from the British Ballroom competitive circuit. The first sample consisted of 50 students that had been previously trained by a prominent dance coach and judge and the second sample were 50 students who had been trained at a rival dance studio. The total score for each student was calculated to find a mean for each group.

If ₁and ₂ represent the mean dance scores for each group respectively for self-coach vs. rival coached, which of the following is the appropriate pair of hypotheses in this case?

H0:12=0Ha:12<0H0:1-2=0Ha:1-2<0

H0:1=2Ha:1<2H0:1=2Ha:1<2

H0:1=2Ha:1>2H0:1=2Ha:1>2

H0:1<2Ha:1=2H0:1<2Ha:1=2

None of the above

Question 5 of 5

Question 5

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Analyses were run. The following is the (edited) output for the test:

Hypothesis Test Results

₁: Dance Scores: Coached Previously

₂: Dance Scores: Not Coached Previously

Difference	Sample Mean	Std. Err.	DF	T-Stat	P-Value
1 - 2	.11	0.67168	94.155815	0.163767	0.4351

From the output we learn that:

the data do not provide sufficient evidence to reject H₀; thus, we cannot conclude that the mean score of the students coached is higher than that of students never coached.

the data provide sufficient evidence reject the H₀; thus, we cannot conclude that the mean score of the students coached is higher than that of students never coached.

the data do not provide sufficient evidence reject the H₀; thus, we can conclude that the mean score of the students coached is higher than that of students never coached.

the data provide sufficient evidence to reject H₀; thus, we can conclude that the mean score of the students coached is higher than that of students never coached.