Test 6 Practice Test #3_____ 1. What is the point estimate of the population mean ?(A) 0 (B) Z (C) ¯X (D) t (E) s2. In the current chapter covering statistical inferences on means, all of the procedures we have discussed (samplingdistributions, confidence intervals and statistical tests) have involved two assumptions. What are those two assumptions?In the fall of 2011 VCU launched Spit for Science: the VCU Student Survey (www.spit4science.vcu.edu). This project isfollowing the 2011 VCU freshman class across its college years, and will also enroll new VCU freshman classes over thenext few years. The goal of the project is to understand how genetic and environmental factors come together toinfluence a variety of health-related outcomes in the VCU undergraduate population.3. Since Spit for Science data is only collected for students in the 2011 VCU freshman class, most of them are similar inage. Suppose that when the project began the ages (in months) of all students in the 2011 VCU freshman class are skewedslightly to the right with a mean of 219 months and a standard deviation of 8 months. If a simple random sample of 44students in the 2011 VCU freshman class is selected and the age in months of each at the beginning of the programdetermined, describe completely the sampling distribution of ¯X, the resulting mean age of this sample of 44 studentsin the 2011 VCU freshman class.4. Data was collected on multiple days, and of interest is to estimate = the mean number of students from which data hasbeen collected each day since the project began. For this problem only, assume that the standard deviation of the numberof students from which data has been collected each day since the project began for all days is 14.7 students. What is theminimum number of days that would need to be selected for data collection to allow the calculation of a 99% confidenceinterval with margin of error no greater than 5.5 students? Please circle your final answer.5. A simple random sample of 50 days was selected and the number of students from which data was collected each day wasrecorded. The mean number of students per day for this sample of 50 days was 21.4 students with a standard deviation of5.2 students, and the distribution is skewed to the right. If appropriate, use this information to calculate and interpret a99% confidence interval for the mean number of students from which data has been collected each day since the projectbegan._____ 6. In question 5 a confidence interval was computed based on a sample of 50 days. If the number of days in thesample were decreased to 30, what impact would this have on the margin of error and width of the confidenceinterval?(A) The margin of error would increase and the width would decrease.(B) The margin of error would decrease and the width would increase.(C) Neither the margin of error nor the width would be affected.(D) Both the margin of error and the width would increase.(E) Both the margin of error and the width would decrease.7. Of interest is to determine the mean number of minutes that all students in the 2011 VCU freshman class have spentdiscussing the Spit for Science project, and since data for all students in the 2011 VCU freshman class is not availablethis mean cannot be determined. It is conjectured that the mean number of minutes that all students in the 2011 VCUfreshman class have spent discussing the Spit for Science project is 180 minutes, and researchers want to test this versusthe alternative that the mean number of minutes is different from 180 minutes. State the appropriate null and alternativehypotheses that should be tested.8. Consider the information and hypotheses specified in question 7. It is known that some students have spent much timediscussing the Spit for Science project, and hence the distribution of time spent discussing is heavily skewed to the right.Also assume that the standard deviation of the time spent discussing the Spit for Science project by all students in the2011 VCU freshman class is 87 minutes. To test the hypotheses in question 7, data for 22 students from the 2011 VCUfreshman class enrolled in a STAT 210 class was recorded. The mean number of minutes the students in the sample spentdiscussing the Spit for Science project was 203 minutes. If appropriate, use this information to test the hypotheses statedin question 7 at the = .05 level of significance.9. Consider the three statements below. Draw a circle around any (if any) and all that are valid statistical hypotheses.H0: = 19.2 HA: > 97 HA: ¯X ≠1610. The “Freshman fifteen” refers to an amount (somewhat arbitrarily set at fifteen pounds) of weight often gainedduring a student's first year at college. With the population being all students in the 2011 VCU freshman class, ofinterest is to test to see if the mean weight gain for this population of students is 15 pounds versus a conjecture supportedby research at Ohio State University that the mean weight gain for all students in the 2011 VCU freshman class is actuallyless than 15 pounds. State the appropriate null and alternative hypotheses that should be tested.11. Consider the information and hypotheses specified in question 10. A simple random sample of 61 students in the 2011VCU freshman class was selected, and the weight gain from when they arrived on campus in August 2011 until March31, 2012 was recorded for each. If a student has lost weight, the weight gain was recorded as a negative number. Themean weight gain for this sample of 61 students was 9.7 pounds, with a standard deviation of 12.3 pounds. If appropriate,use this information to test the hypotheses stated in question 10 at the = .10 level of significance.Test 6 Practice Test #3 Solutions1. C – The sample mean ´X is the point estimate of the population mean µ.2. The assumptions are:1. Simple random sample2. Normal population or large enough sample for the Central Limit Theorem to apply.3. We have a simple random sample, and the sample size is large enough to apply the Central Limit Theorem (n = 44 >15). So μ´X=μ = 219; σ´X=σ√n=8√44= ¿ 1.206; since the CLT applies, the shape is normal. Hence¯X ~ N(219, 1.206).4. n =(Z¿σm)2=(2.576(14.7)5.5)2 = (6.885)2 = 47.4; Round up to 48 days.5. We have a simple random sample, and the sample size is large enough for the Central Limit Theorem to apply (n = 50> 15), so the assumptions are
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