1You run a gizmo factory. Over the past year the average output of the factory is 480 gizmos and the

1You run a gizmo factory. Over the past year the average output of the factory is 480 gizmos and the

1You run a gizmo factory. Over the past year the average output of the factory is 480 gizmos and the standard deviation of that output is 10 units per day. In repeated random samples of size n = 40 days, the expected value of sample mean is:a12b480c76d482In the previous question the variance of the sample mean is.a10b2.5c15.8d1.583Using the population mean and standard deviation in question 1, what is the probability that the mean of a random sample of n = 40 days exceeds 482 gizmos: a0.1020b0.0571c0.0384d0.02874Using the population mean and standard deviation in question 1, in repeated samples of size n = 40, 95% of sample means fall within that fall within ¤ gizmos from the population means. a3.9b3.5c3.1d2.75In the population of employees in your company, 35 percent contribute to the annual United Way campaign. You plan to take a sample of size n = 200 and calculate the sample proportion who contribute to United Way.The expected value of the sample proportion is:a0.035b0.25c0.35d0.406In the previous question, the standard error of the sample proportion is:a0.0473b0.0415c0.0398d0.03377In the previous question, 90 percent of sample proportions from samples of size n = 100 deviate from the population proportion of 0.35 by no more that ¤ (or percentage points).a0.036(3.6 percentage points.)b0.042(4.2 percentage points.)c0.048(4.8 percentage points.)d0.055(5.5 percentage points.)8There is a population of 10 families in a small neighborhood. You plan to take a random sample of 4 families (without replacement). How many samples of size n = 4 are possible.a40b80c210d4009The expressionmeans:aIn repeated sampling the probability that population mean is within ¤1.96??/??n from x?? is 0.95.bIn repeated sampling the probability that x?? is within ¤1.96??/??n from the population mean is 0.95.c95% of sample means deviate from the population mean by no more than 1.96??/??n in either direction.dBoth b and c are correct.10The population proportion of Americans with diabetes is 9 percent (? = 0.09). In repeated random samples of n = 800 Americans, 90% of sample proportions of people with diabetes would fall within ¤ from ?.a0.01(1.0 percentage point)b0.017(1.7 percentage points)c0.029(2.9 percentage points)d0.036(3.6 percentage points)11As part of a statistics assignment in October to estimate the percentage of voters who would vote for a mayoral candidate, each of 500 students collects his or her own random sample of likely voters. There are 400 voters in each students random sample. Each student then constructs a 95 percent confidence interval for the population proportion who will vote for the candidate using his or her own random sample. Considering the 500 intervals constructed by the students, the expected number of intervals that will contain the population proportion who will vote for that candidate will be approximately:a380b400c475d50012Suppose the sample proportion of one of the students in the previous question, Beths sample, is p?? = 0.46. Beths 95% interval estimate of proportion of the population of voters voting for the candidate is:a[0.41 , 0.51]b[0.40 , 0.52]c[0.39 , 0.53]d[0.38 , 0.54]13It is estimated that 80% of Americans go out to eat at least once per week, with a margin of error of 0.04 (for 95% confidence). A 95% confidence interval for the population proportion of Americans who go out to eat once per week or more is:a[0.722, 0.878]b[0.760, 0.840]c[0.771, 0.829]d[0.798, 0.802]14You are the manager of a political campaign. You think that the population proportion of voters who will vote for your candidate is 0.50 (use this for a planning value). Your candidate wants to know what proportion of the population will vote for her. Your candidate wants to know this with a margin-of-error of ¤ 0.01 (at 95% confidence). How big of a sample of voters should you take?a1499b5037c8888d960415If your candidate changes her mind and now wants a margin-of-error of ¤ 0.03 (but still 95% confidence), you will:ahave to select a smaller samplebbe able to use the same samplechave to select a larger sampledthe margin-of-error does not have anything to do with the sample size16You run a bank and want to estimate the banks average number of customers per day (the population is all the days you are open for business in a year). You take a random sample of 10 days and record the numbers of customers on those days. The sample data is shown below. What is a 95% confidence interval for the banks average number of customers per day?450470430420490440460420500420a[433, 467]b[430, 470]c[427, 473]d[424, 476]


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