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In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelors
In a study of government financial aid for college students, it becomes necessary to estimate the percentage of full-time college students who earn a bachelors degree in four years or less. Find the sample size needed to estimate that percentage. Use a 0.04 margin of error and use a condence level of 95%. Complete parts (a) through (c) below. a. Assume that nothing is known about the percentage to be estimated. n = (Round up to the nearest integer.) You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you survey? Assume that you want to be 95% condent that the sample percentage is within 3.5 percentage points of the true population percentage. Complete parts (a) and (b) below. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. n = (Round up to the nearest integer.) Assume that we want to construct a condence interval. Do one of the following, as appropriate: (a) nd the critical value tuiZ' (b) nd the critical value Zn\Assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) find the critical value to /2, (b) find the critical value Zo /2, or (c) state that neither the normal distribution nor the t distribution applies. Here are summary statistics for randomly selected weights of newborn girls: n = 164, x = 30.3 hg, s = 7.6 hg. The confidence level is 99%. . . . Select the correct choice below and, if necessary, fill in the answer box to complete your choice. OA. to/2= (Round to two decimal places as needed.) OB. Za/2 = (Round to two decimal places as needed.) O C. Neither the normal distribution nor the t distribution applies.Here are summary statistics for randomly selected weights of newborn girls: n = 244, x = 30.2 hg, s = 6.2 hg. Construct a confidence interval estimate of the mean. Use a 95% confidence level. Are these results very different from the confidence interval 29.2 hg
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