finding percentages using confidence intervals, statistics assignment help

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OboWbarf84

Mathematics

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The first question involves finding percentages using confidence intervals.

The second question involves looking at the question right above it in bold. You would use 11.28 to answer to the second question.

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1.) For a random sample of 100 households in a large metropolitan area, the number of households in which at least one adult is currently unemployed and seeking a full-time job is 12. Estimate the percentage of households in the area in which at least one adult is unemployed, using a 95 percent confidence interval. 11.28) The sponsor of a television special expected that at least 40 percent of the viewing audience would watch the show in a particular metropolitan area. For a random sample of 100 households with television sets turned on, 30 of them are found to be watching the special. Designating the sponsor’s assumption the benefit of doubt, can it be rejected as being true for all viewers in the metropolitan area at the (a) 10 percent and (b) 5 percent level of significance? 2.) For Problem 11.28, suppose the sponsor specifies that as the result of the study the probability of rejecting a true claim should be no larger than P = 0.02, and the probability of accepting the claim given that the percentage viewing the program is actually 30 percent or less should be no larger than P= 0.05. What sample size is required in the study, as a minimum, to satisfy this requirement?
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Explanation & Answer

Here it is! I hoep you love it.

1.) For a random sample of 100 households in a large metropolitan area, the number of
households in which at least one adult is currently unemployed and seeking a full-time
job is 12. Estimate the percentage of households in the area in which at least one adult is
unemployed, using a 95 percent confidence interval.
Here it is .12*100 or 12% that are currently unemployed in this situation. Using the 95%
confidence interval it is possible to take .95+.12=1.07...


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