(use = 0.05) for a) and b) a) A poll done or Newsweek found that 3% o amer cans

Question

(use = 0.05) for a) and b) a) A poll done or Newsweek found that 3% o amer cans have seen or sensed the presence o an angel A contingent doubts t at the percent s real y at hi . t conducts its o n survey utof Sinner cans un ve onl e had seen or sensed the presence of an angel. As a result of the contingents survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. b) The average work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's shorter. She asks 10 engineering friends in start-ups for the lengths of their average work weeks, Based on the results that follow, should she count on the average work week to be shorter than 60 hours? Data (length of average work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55

Explanation / Answer

(a) Pr(sensed or felt the presence of angel) = 0.13

Here sample proportion p^ = 2/76 = 0.0263

Here,

H0 : p = 0.13

Ha : p < 0.13

standard error of the proprtion se0 = sqrt [0.13 * 0.87/ 76] = 0.0386

Test statistic  

Z = (p^ - 0.13)/ se0  = (0.0263 - 0.13)/ 0.0386 = -2.6865

Here p -value = Pr (Z < -2.6865) = 0.0036 < 0.05

so we shall reject the null hypothesis and can claim that Newsweek poll is not correct.

Reasons that two polls give different results.

(i) Here sample population for the newsweek poll and the contingent may be different. As less literate people tend to believe in angels then literate people so there may be chance that sample population may be different,

(ii) Here may be there is voluntary biases in sample. As newsweek is a popular megazine, people tend to give there preference in it.

(iii) There may not be a random sampling by the contigent. It may have taken a random sample from a limited population.

(b)

H0 : = 60 weeks

Ha : < 60 weeks

sample mean of average work weeks x = 57 weeks

sample standard deviation s = 7.1492 weeks

Here,

stadard error of sample mean = s/ sqrt (n) = 7.1492/ sqrt(10) = 2.261 weeks

Test statistic

t = (x -H)/se0 = (57 - 60)/ 2.261 = -1.327

so critical value of t

tcritical = t9,0.05  = 2.2622

so l t l < tcritical  so we shall not reject the null hypothesis and can conclude that average work week is 60 weeks.

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