In 1997 a survey 1500 randomly selected adults showed that 39% of them had taken

Question

In 1997 a survey 1500 randomly selected adults showed that 39% of them had taken more vacations this year than last year. Construct a 95% confidence interval estimate of the population proportion of adults that had taken more vacations this year than last year. State the Margin of Error, Best point estimate and Include the written statement In 2014, six percent of the cars sold had a manual transmission. A random sample of college students who cars owned cars revealed the following: out of 122 cars, 26 had manual transmissions. Estimate the proportion of college students who drive cars with manual transmissions with 90% confidence. State the Margin of Error, Best point estimate and Include the written statement

Explanation / Answer

(6)
n = 1500
p = 0.39

Standard Error SE = sqrt(p*(1-p)/n) = sqrt(0.39*(1-0.39)/1500) = 0.0126

For 95% CI, z-value = 1.96

Margin of Error, ME = z*SE = 1.96*0.0126 = 0.0247

Confidence Interval = (p-ME, p+ME) = (0.39-0.0247, 0.39+0.0247) = (0.3653, 0.4147)

This means for any selected random sample of size 1500, there is 95% confidence that proportion of adults that had taken more vacations this year than last year lies in the interval (0.3653, 0.4147)

(7)
n = 122
p = 26/122 = 0.2131

Standard Error SE = sqrt(p*(1-p)/n) = sqrt(0.2131*(1-0.2131)/122) = 0.0371

For 90% CI, z-value = 1.64

Margin of Error, ME = z*SE = 1.64*0.0371 = 0.0608

Confidence Interval = (p-ME, p+ME) = (0.2131-0.0608, 0.2131+0.0608) = (0.1523, 0.2739)

This means for any selected random sample of size 122, there is 90% confidence that proportion of college students who drive cars with manual transmission lies in the interval (0.1523, 0.2739)

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