A graduate student wishes to know the proportion of U.S adults who speak two or

Question

A graduate student wishes to know the proportion of U.S adults who speak two or more languages. He surveys 428 US, adults and finds that 107 speak two or more languages. Construct a 99% confidence interval to estimate the proportion of all U.S. adults that speak two or more languages The Montana State Education Commission wants to estimate the percentage of tenth-grade students who read at or below the eighth-grade level. How large a sample should be selected in order to estimate this proportion with a 98% confidence level and a margin of error of at most 2%? (Assume no preliminary estimate of p-hat is available) A toy company wants to know the average number of new toys bought for children each year

Explanation / Answer

(16)
p=107/428 = 0.25
SE = sqrt(p*(1-p)/n) = sqrt(0.25*(1-0.25)/428) = 0.0209

For 99%, z-value = 2.57
lower limit = 0.25 - 2.57*0.0209 = 0.1963
upper limit = 0.25 + 2.57*0.0209 = 0.3037

(17)
Margin of error = z * SE
for 98% CI, z = 2.33

0.02 = 2.33 * SE
SE = 0.00858369

lets assume phat = 0.5
sqrt(0.5*0.5/n) = 0.00858369
n = (0.5*0.5/0.00858369)^2
n = 848.2658201

Hence sample should be selected of 848.

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