-LUWcbnyklFxR8dx153bzRIQlaqToGEm&effectiveUsersmaldhafeeriBuseremaldh; MATHEMATI

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-LUWcbnyklFxR8dx153bzRIQlaqToGEm&effectiveUsersmaldhafeeriBuseremaldh; MATHEMATICAL ASSOCIATION OF AMERICA webwork, m366ryu I set6chap7 13 Set6Chap7: Problem 3 Previous Problem List Next point) in developing an interval estimate for a population mean, the interval estimate was 62.84 to 69 46. The population standard deviation was assumed to be 6.50, and a sample of observations was used. The mean of the sample was 100 66.15 D. 13 24 The minimum sample see needed to a population mean within 2 units with a 95% confidence when the standard population equals B is estimate C, 61

Explanation / Answer

1.We know that Confidence Interval is Mean+/-Margin of Error

Here upper bound is mean+error=69.46 and

lower bound is mean-error=62.84

Soving this two equation we get 2 mean=69.46+62.84

Hence 2mean=132.3

So mean=66.15. B is answer

2. Here margin of error =2, for 95% CI z value is 1.96, as (-1.96<z<1.96)=0.95 and sd=8

Now we know formula for Margin of Error=E=z*sd/sqrt(n)

So n=(z*sd/E)^2=(1.96*8/2)^2=61.4=61 so C is answer

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