Question1 Inthe2006GeneralSocialSurvey,respondentswereaskedwhethertheyfavored(success)or opposed the death penalty for people convicted of murder. The results found that x = 1885 with n = 2815. (a)...

Question1 Inthe2006GeneralSocialSurvey,respondentswereaskedwhethertheyfavored(success)or opposed the death penalty for people convicted of murder. The results found that x = 1885 with n = 2815. (a) What is the sample proportion, pˆ?

(b) Can you conclude that more than half of all American adults are in favor? Why? (c) Find a 95% confidence interval for the proportion of American adults who opposed the death penalty. Interpret.

Question2

Accordingtoaunionagreement,themeanincomeforallsenior-levelassembly-lineworkers in a large company equals $500 per week. A representative of a women’s group decides to analyze whether the mean income μ for female employees matches this norm. For a random sample of nine employees, Y ̄ = 410 and s = 90.

a. Test whether the mean income of female employees differed from $500 per week. Include assumptions, hypotheses, test statistic, and p-value. Interpret the results.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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In the survey being conducted to see whether Americans are in favor of the death penalty or not, 2815 people make up the sample size. In a survey, the accuracy of the result is reduced by the size of the sample. The sensitivity of the survey conducted in estimating the correct outcome can be arrived at as follows.

Take the desired sensitivity, here it is 95% or 0.95. Subtract this from 1, 1 - 0.95 = 0.05. Multiply the two values, 0.95*0.05 = 0.0475. Divide this by the sample size, here it is 2815, `0.0475/2815 ~~ 1.687*10^-5` . Take the square root of this value, `sqrt(0.0475/2815 ) ~~ 4.1077*10^-3` . For the 95% confidence interval the corresponding value in normal distribution tables is 1.96.

`4.1077*10^-3*1.96 = 8.05*10^-3`

The value obtained above added and subtracted from 0.95 provides the 95% confidence interval relevant to the survey outcome. It is 0.9580512 to 0.9419. The 95% confidence interval for the survey is 0.9580 to 0.9419

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