Out of a sample of 327 Americans, 245 said they had no interest in professional soccer. (Data simulated from Carey & Kereslidze, 2007) A 95% confidence interval for the
proportion of Americans who have no interest in professional soccer is 0.70 to 0.80.
Suppose that another sample of 784 Americans was taken and asked the same question.
How would the width of the new confidence interval compare to the width of the
confidence interval based on the 327 women in the armed forces?
The width of the confidence interval based on the 784 women would be narrower than the
confidence interval based on the 327 women.
The width of the confidence interval based on the 784 women would be wider than the
confidence interval based on the 327 women.
The width of the confidence interval based on the 784 women would be the same as the
confidence interval based on the 327 women.
O No answer text provided.

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

The margin of error is:

From this, we have that the margin of error, and also the width, is inversely proportional to the sample size, that is, a larger sample size leads to a smaller margin of error and a narrower interval.

How would the width of the new confidence interval compare to the width of the  confidence interval based on the 327 women in the armed forces?

New interval: 784

Old interval: 327

Sample size increased, so the new interval will be narrower, and the correct answer is:

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

.19875994384

step-by-step explanation:

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hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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step-by-step explanation: