MATH 225N Week 8 Final Exam (Version 1) (1).pdf, Exams of Nursing

Download MATH 225N Week 8 Final Exam (Version 1) (1).pdf and more Exams Nursing in PDF only on Docsity! MATH 225N Week 8 Final Exam (Version 1) Question 1 · 1/1 points A fitness center claims that the mean amount of time that a person spends at the gym per visit is 33minutes. 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: μ<33 H0: μ≤33; Ha: μ>33 Answer Explanation Correct answer: H0: μ=33; Ha: μ≠33 Let the parameter μbe 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 or equal to (≤). Therefore, in this case, the null hypothesis H0is μ=33. The alternative hypothesis is contradictory to the null hypothesis, so Hais μ≠33. • MATH 225N Week 8 Final Exam (Version 1) • • • Question 2 · 1/1 points The answer choices below represent different hypothesis tests. Which of the choices are right- tailed tests? Select all correct answers. That is correct! • H0:X≥17.1, Ha:X<17.1 • • H0:X=14.4, Ha:X≠14.4 • • H0:X≤3.8, Ha:X>3.8 • MATH 225N Week 8 Final Exam (Version 1) structures were built without permits when, in fact, more than 15%of the structures were built without permits. • • • • Question 4 · 1/1 points Suppose a chef claims that her meatball weight is less than 4ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10%s ign if ica nce level, to persuade them. She cooks 14meatba lls. The mean weight of the sample meatballs is 3.7o u n c e s . The chef knows from experience that the standard deviation for her meatball weight is 0.5o u n c e s . • H0: μ≥4; Ha: μ<4 • α=0.1(s ignific a nce level) What is the test statistic (z- score) of this one- mean hypothesis test, rounded to two decimal places? That is correct! Test statistic = minus 2 point 2 4$$ Test statistic = minus 2 point 2 4 - correct Answer Explanation Correct answers: MATH 225N Week 8 Final Exam (Version 1) • Test statistic = minus 2 point 2 4 $\text{Test statistic = }-2.24$ • The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x̄=3.7. The sample the chef uses is 14meatba lls , so n=14. She knows the standard deviation of the meatballs, σ=0.5. Lastly, the chef is comparing the population mean weight to 4ounces. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x̄−μ0σn√=3.7−40.514√≈−0.30.134≈−2.24 So, the test statistic for this hypothesis test is z0=−2.4 • • • • Question 5 · 1/1 points What is the p-value of a right-tailed one- mean hypothesis test, with a test statistic of z0=1.74? (Do not round your answer; compute your answer using a value from the table below.) z1.51.61.71.81.90.000.9330.9450.9550.9640.9710.010.9340.9460.9560.9650. 9720.020.9360.9470.9570.9660.9730.030.9370.9480.9580.9660.9730.040.93 80.9490.9590.9670.9740.050.9390.9510.9600.9680.9740.060.9410.9520.961 0.9690.9750.070.9420.9530.9620.9690.9760.080.9430.9540.9620.9700.9760 .090.9440.9540.9630.9710.977 That is correct! MATH 225N Week 8 Final Exam (Version 1) 0 point 0 4 1$$ 0 point 0 4 1 - correct Answer Explanation Correct answers: • 0 point 0 4 1 $0.041$ • The p-value is the probability of an observed value of z=1.74or greater if the null hypothesis is true, because this hypothesis test is right-tailed. This probability is equal to the area under the Standard Normal curve to the right of z=1.74. A standard normal curve with two points labeled on the horizontal axis. The mean is labeled at 0.00 and an observed value of 1.74 is labeled. The area under the curve and to the right of the observed value is shaded. Using the Standard Normal Table, we can see that the p-value is equal to 0.95, which is the area to the left of z=1.74. (Standard Normal Tables give areas to the left.) So, the p-value we're looking for is p=1−0.959=0.041. • • • • MATH 225N Week 8 Final Exam (Version 1) 34$$ 34 - correct Answer Explanation Correct answers: • 34 $34$ • By using the known totals along the rows and columns you can fill in the rest of the contingency table. For example, looking at the second row in the table, we know that 33a d d e d to the unknown number in the middle is 67, so that unknown number is 34. Continuing in this way, we can fill in the entire table: StudentsplaysportsdonotplaysportsTotalplayaninstrument273360donotplayani nstrument353469Total6267129 From this, we can see that the number of students who both do not play sports and do not play an instrument is 34. • • • • Question 46 · 1/1 points The answer choices below represent different hypothesis tests. Which of the choices are left- tailed tests? Select all correct answers. That is correct! MATH 225N Week 8 Final Exam (Version 1) • H0:X=17.3, Ha:X≠17.3 • • H0:X≥19.7, Ha:X<19.7 • • H0:X≥11.2, Ha:X<11.2 • • H0:X=13.2, Ha:X≠13.2 • • H0:X=17.8, Ha:X≠17.8 • Answer Explanation Correct answer: MATH 225N Week 8 Final Exam (Version 1) H0:X≥19.7, Ha:X<19.7 H0:X≥11.2, Ha:X<11.2 Remember the forms of the hypothesis tests. • Right-tailed: H0:X≤X0, Ha:X>X0. • Left-tailed: H0:X≥X0, Ha:X<X0. • Two-tailed: H0:X=X0, Ha:X≠X0. So in this case, the left-tailed tests are: • H0:X≥11.2, Ha:X<11.2 • H0:X≥19.7, Ha:X<19.7 • • • • Question 47 · 1/1 points Assume the null hypothesis, H0, is: Jacob earns enough money to afford a luxury apartment. Find the Type I error in this scenario. That is correct! Jacob thinks he does not earn enough money to afford the luxury apartment when, in fact, he does. Jacob thinks he does not earn enough money to afford the luxury apartment when, in fact, he does not. MATH 225N Week 8 Final Exam (Version 1) The means of Aand Bare equal. • • Ahas the larger standard deviation. • • Bhas the larger standard deviation. • • The standard deviations of Aand Bare equal. • Answer Explanation Correct answer: The means of Aand Bare equal. Ahas the larger standard deviation. Remember that the mean of a normal distribution is the x- value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean. Because Aand Bare centered at the same point, their means are equal. Remember that the standard deviation tells how spread out the normal distribution is. So a high standard deviation means the graph will be short and spread out. A low standard deviation means the graph will be tall and skinny. MATH 225N Week 8 Final Exam (Version 1) Because Ais shorter and more spread out than B, we find that Ahas the larger standard deviation. • • • • Question 49 · 1/1 points Hugo averages 62w o r d s per minute on a typing test with a standard deviation of 8words per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X=t he number of words per minute on a typing test. Then, X∼N(62,8). Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is . This z-score tells you that x=56is standard deviations to the (right/left) of the mean, . Correctly fill in the blanks in the statement above. That is correct! Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is 0.75. This z-score tells you that x=56is 0.75s t a nd a r d deviations to the right of the mean, 62. MATH 225N Week 8 Final Exam (Version 1) Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is −0.75. This z-score tells you that x=56is 0.75s t a nd a r d deviations to the left of the mean, 62. Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is 0.545. This z-score tells you that x=56is 0.545standard deviations to the right of the mean, 62. Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is −0.545. This z-score tells you that x=56is 0.545standard deviations to the left of the mean, 62. Answer Explanation Correct answer: Suppose Hugo types 56w o r d s per minute in a typing test on Wednesday. The z-score when x=56is −0.75. This z-score tells you that x=56is 0.75s t a nd a r d deviations to the left of the mean, 62. The z-score can be found using the formula z=x−μσ=56−628=−68≈−0.75 A negative value of zmea ns that that the value is below (or to the left of) the mean, which was given in the problem as μ=62words per minute in a typing test. The z-score tells you how many standard deviations the value xis above (to the right of) or below (to the left of) the mean, μ. So, typing 56w o r d s per minute is 0.75standard deviations away from the mean. •