Chapter 9, Problem 9.63RE

a)

To determine

To explain why we cannot say that 37% of the adults would say that football is their favorite sport to watch on television.

Answer to Problem 9.63RE

Because of sampling variability.

Explanation of Solution

Given:

n=1000

Sample proportion:

  $\widehat{p}=37\%=0.37$

Confidence level = 95% = 0.95

First need to understand about sample and population.

Population: It is the set of all the possible individuals possessing the characteristic of interest in a study.

Sample: it is the subset of population on which, the actual study performed and from which, the necessary measurements are taken.

The estimated percentage 37% is based on sample. We know that does not mean it should apply on true percentage of population. There is sampling variability present. That’s why we cannot say that 37% of the adults would say that football is their favorite sport to watch on television.

b)

To determine

To find the 95% confidence interval for p.

Answer to Problem 9.63RE

The 95% confidence interval is, (0.3401, 0.3999)

Explanation of Solution

Given:

n=1000

Sample proportion:

  $\widehat{p}=37\%=0.37$

Confidence level = 95% = 0.95

Formula:

Confidence interval for population proportion is,

  $CI=\widehat{p}\mp Zc\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Calculation:

Now need to find Zc

The level of significance is = a =0.05

Therefore, Zc = Z _a/2 = 1.96 ..Using excel function, =ABS(NORMSINV(0.05/2))

Therefore, the 95% confidence interval for population proportion is,

  $\begin{array}{l}\left(0.37-1.96\sqrt{\frac{0.37(1-0.37)}{1000}},0.37+1.96\sqrt{\frac{0.37(1-0.37)}{1000}}\right)\\ \left(0.3401,0.3999\right)\end{array}$

c)

To determine

To explain whether to agree with margin of error $\pm 3\%$

Answer to Problem 9.63RE

Agree.

Explanation of Solution

Given:

n=1000

Sample proportion:

  $\widehat{p}=37\%=0.37$

Confidence level = 95% = 0.95

Formula:

Margin of error:

  $E=Zc\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}}$

Calculation:

Now need to find Zc

The level of significance is = a =0.05

Therefore, Zc = Z _a/2 = 1.96 ..Using excel function, =ABS(NORMSINV(0.05/2))

Therefore, the margin of error is,

  $E=1.96\sqrt{\frac{0.37(1-0.37)}{1000}}=0.02999\approx 0.03=3\%$

It is approximately 3% hence agree.

d)

To determine

To explain if this poll was taken in December, an exciting part of the football season. whether a similar poll conducted in June might produce different results.

Answer to Problem 9.63RE

Yes

Explanation of Solution

There might be a situation where other sports will dominate television in June compared to December. Therefore, people are more likely to mention the sport that they last watched on television.