# It is reported in USA Today that the average flight cost nationwide is $458.78. You have... ###### Question: It is reported in USA Today that the average flight cost nationwide is$458.78. You have never paid close to that amount and you want to perform a hypothesis test that the true average is actually less than $458.78. The hypotheses for this situation are as follows: Null Hypothesis: μ ≥ 458.78, Alternative Hypothesis: μ < 458.78. You take a random sample of national flight cost information and perform a one sample mean hypothesis test. You observe a p-value of 0.9892. What is the appropriate conclusion? Conclude at the 5% level of significance. Question 12 options:  1) The true average flight cost is greater than or equal to$458.78.
 2) We did not find enough evidence to say the true average flight cost is less than $458.78.  3) We did not find enough evidence to say the true average flight cost is greater than$458.78.
 4) The true average flight cost is significantly less than $458.78.  5) We did not find enough evidence to say a significant difference exists between the true average flight cost and$458.78.

Do sit down restaurant franchises and fast food franchises differ significantly in stock price? Specifically, is the average stock price for sit-down restaurants different from the average stock price for fast food restaurants? If sit down restaurants are considered group 1 and fast food restaurants are group 2, what are the hypotheses of this scenario?

Question 13 options:

 1) HO: μ1 ≥ μ2 HA: μ1 < μ2
 2) HO: μ1 > μ2 HA: μ1 ≤ μ2
 3) HO: μ1 ≤ μ2 HA: μ1 > μ2
 4) HO: μ1 ≠ μ2 HA: μ1 = μ2
 5) HO: μ1 = μ2 HA: μ1 ≠ μ2

It is believed that students who begin studying for final exams a week before the test score differently than students who wait until the night before. Suppose you want to test the hypothesis that students who study one week before score greater than students who study the night before, giving you the following hypotheses: Null Hypothesis: μ1 ≤ μ2, Alternative Hypothesis: μ1 > μ2. A random sample of 47 students who indicated they studied early shows an average score of 88.26 (SD = 3.299) and 28 randomly selected procrastinators had an average score of 89.21 (SD = 5.444). Perform a two independent samples t-test assuming that early studiers are group 1 and procrastinators are group 2. What is the test statistic and p-value of this test? Assume the population standard deviations are the same.

Question 14 options:

 1) Test Statistic: 0.943, P-Value: 0.1744
 2) Test Statistic: 0.943, P-Value: 0.8256
 3) Test Statistic: -0.943, P-Value: 0.8256
 4) Test Statistic: -0.943, P-Value: 0.1744
 5) Test Statistic: -0.943, P-Value: 1.6512

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