Each person in a random sample of 2000 “likely voters” (as defined by a professi

Question

Each person in a random sample of 2000 “likely voters” (as defined by a professional polling organization) was questioned about his or her political views. Of those surveyed, 1308 felt that “the economy's state” was the most urgent national concern. A 99% confidence interval for the proportion p of all likely voters that feel the economy's state is the most urgent national concern is given by (use the plus four confidence interval procedure)

0.624 to 0.663.

0.615 to 0.672.

0.626 to 0.681.

0.606 to 0.680.

Explanation / Answer

Let's write the given information

n = sample size = 2000

Suppose X be a random variable which takes the number of likely voters that feel the economy's state is the most urgent national concern.

So here x = 1308 because out of 2000 likely voters 1308 felt that “the economy's state” was the most urgent national concern.

Using minitab.

Using minitab.

The command for one sample proportion z interval in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 1308

Number of trials = n = 2000

then click on option

Level of confidence in percentage = c = 99%

so put "Confidence level " = 99.0

Alternative = not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

Test and CI for One Proportion

Sample X N Sample p 99% CI
1 1308 2000 0.654000 (0.626, 0.681)

Using the normal approximation.

From the above minitab ouput: The 99% confidence interval for the proportion p of all likely voters that feel the economy's state is the most urgent national concern is (0.626, 0.681).

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