A Wireless News article on July 6, 2008 found that in a random survey of 600 cel

Question

A Wireless News article on July 6, 2008 found that in a random survey of 600 cell phone users, 372 would use a Bluetooth device while driving in order to comply with the law. a Construct a 99% confidence interval for the proportion of all cell phone users who will use Bluetooth technology while driving to comply with the law. b One cell phone manufacturer wishes to estimate the proportion of cell phone users that will use Bluetooth technology while driving. To do so, they will construct a 99% confidence interval. If they want the interval to be no wider than 5 percentage points, how many cell phone users must they sample if they use the Wireless News article to estimate p in their calculation?

Explanation / Answer

Answer) n = 600 x = 372

              The sample proportion p = x / n = 372 / 600

                                                   p = 0.62

(a) We need to determine 99% Confidence Interval

The formula for the confidence interval is

                             p +/- Z* sqrt(pq/n)

              0.62  +/- 2.58 * sqrt(0.62(1-0.62) / 600)

             0.62 +/- 0.05113

Hence, 99% confidence interval is (0.56887, 0.67113)

(b) We need to determine the sample size here

Margin of Error must not be wider than 5% points

So, M.E = 5% = 0.05

M.E = Z * sqrt(p(1-p) / n)

So, Margin of Error must not be wider than 5% points

0.05 > 2.58* sqrt(0.62(1-0.62) / n)

0.05 > 1.2523/sqrt(n)

sqrt(n) < 1.2523 / 0.05

sqrt(n) < 25.046

n < 627.30

Hence, the sample size must be lesser than 628 to get the interval no wider than 5 percentage points

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