MATH 225N Statistics Final Exam

MATH 225N Statistics Final Exam 1/1 POINTS A fitness center claims that the mean amount of time that a person spends at the gymper visit is 33 minutes. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter μ. That is correct! H0: μ≠33; Ha: μ=33 H0: μ=33; Ha: μ≠33 H0: μ≥33; Ha: μ33 H0: μ≤33; Ha: μ33 Answer Explanation Correct answer: Let the parameter μ be used to represent the mean. The null hypothesis is always stated with some form of equality: equal (=), greater than or equal to (≥), or less than orequal to (≤). Therefore, in this case, the null hypothesis H0 is μ=33. The alternative hypothesis is contradictory to the null hypothesis, so Ha is μ≠33. QUESTION 21/1 POINTS The answer choices below represent different hypothesis tests. Which of the choicesare right-tailed tests? Select all correct answers. That is correct! H0:X≥17.1, Ha:X17.1 H0: μ=33; Ha: μ≠33       H0:X=14.4, Ha:X≠14.4 H0:X≤3.8, Ha:X3.8 H0:X≤7.4, Ha:X7.4 H0:X=3.3, Ha:X≠3.3 Answer Explanation Correct answer: Remember the forms of the hypothesis tests.  Right-tailed: H0:X≤X0, Ha:XX0.  Left-tailed: H0:X≥X0, Ha:XX0.  Two-tailed: H0:X=X0, Ha:X≠X0. So in this case, the right-tailed tests are:  H0:X≤7.4, Ha:X7.4  H0:X≤3.8, Ha:X3.8 QUESTION 3 H0:X≤3.8, Ha:X3.8 H0:X≤7.4, Ha:X7.4   0/1 POINTS Find the Type II error given that the null hypothesis, H0, is: a building inspector claimsthat no more than 15% of structures in the county were built without permits. That's not right. The building inspector thinks that no more than 15% of the structures in the county werebuilt without permits when, in fact, no more than 15% of the structures really were built without permits. The building inspector thinks that more than 15% of the structures in the county werebuilt without permits when, in fact, more than 15% of the structures really were built without permits. The building inspector thinks that more than 15% of the structures in the county werebuilt without permits when, in fact, at most 15% of the structures were built without permits. The building inspector thinks that no more than 15% of the structures in the county werebuilt without permits when, in fact, more than 15% of the structures were built without permits. Answer Explanation Correct answer: A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. Inthis case, the Type II error is when the building inspector thinks that no more than 15% of the structures were built without permits when, in fact, more than 15% ofthe structures were built without permits. Your answer: Content attribution- Opens a dialog The building inspector thinks that more than 15% of the structures in the county were built without permits when, in fact, at most 15% of the structures were built without permits. The building inspector thinks that no more than 15% of the structures in the county were built without permits when, in fact, more than 15% of the structures were built without permits. QUESTION 41/1 POINTS Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10% significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experiencethat the standard deviation for her meatball weight is 0.5 ounces. H0: μ≥4; Ha: μ4 α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to twodecimal places? That is correct! $$Test statistic = −2.24 Answer Explanation Correct answers:  $text{Test statistic = }-2.24$Test statistic = −2.24 The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x¯=3.7. The sample the chef uses is 14 meatballs, so n=14. She knowsthe standard deviation of the meatballs, σ=0.5. Lastly, the chef is comparing the population mean weight to 4 ounces. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=3.7−40.514√≈−0.30.134≈−2.24 So, the test statistic for this hypothesis test is z0=−2.24.