Mathematics, 06.06.2020 02:01 potato3999

A recent survey collected information on television viewing habits from a random sample of 1,000 people in the United States. Of those sampled, 37 percent indicated that their favorite sport to watch on television was American football. Construct and interpret a 95 percent confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football. Based on your answer to the previous problem, is it reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

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Answer from: bmartinez26

The 95% Confidence Interval we will found for given case is:

(0.34, 0.40)

It is Not reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

Given that:Total size of sample = n = 1000Percent of people in sample indicating that their favorite sport on TV is to watch American football = p = 37% = 0.37 probability.To find:

95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football.

Finding the Confidence Interval:

When there is n sample size and the probability of success is p with level of significance $\alpha$ calculated as:

$CI = p \pm z_0 \sqrt{\dfrac{p(1-p)}{n}$

Where z_0 is the z score $1- \dfrac{\alpha}{2}$ .

1 - $\alpha$ = 0.95 ( since it is given to find 95% CI)

Thus, $\alpha$ = 0.05

The z score at p value = 1 - $\alpha$ /2 = 1 -0.025 = 0.975 is 1.96

Thus we have:

$CI = p \pm z_0 \sqrt{\dfrac{p(1-p)}{n}}\\\\CI = 0.37 \pm 1.96 \sqrt{\dfrac{0.37 \times0.63 }{1000}}\\\\CI = 0.37 \pm 1.96 \times 0.0152 = 0.37 \pm 0.03\\$

Lower limit of CI = 0.34 and Upper limit of CI = 0.40.

The case to decide to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football can be solved as:

33% = 0.33 probability of actually liking watching american probability.

Since 0.33 is lower than the lower limit of CI for 0.37 probability, thus we reject the null hypothesis that both 0.37 and 0.33 are approximately same, and thus decide that it is false that 0.33 percent believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football.

Thus it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football.

Learn more about Confidence interval here:

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Answer from: sarahelrod9

The 95% confidence interval for the proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is (0.34, 0.40). This means that we are 95% sure that the true proportion of all people in the United States who would indicate that their favorite sport to watch on television is American football is between (0.34 and 0.40).

The lower bound of the confidene interval is higher than 0.33. So it is not reasonable to believe that 33% is the actual percent of people in the United States whose favorite sport to watch on television is American football, this estimate should be higher than 33%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1000, \pi = 0.37$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 - 1.96\sqrt{\frac{0.37*0.63}{1000}} = 0.34$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.37 + 1.96\sqrt{\frac{0.37*0.63}{1000}} = 0.40$

Based on your answer to the previous problem, is it reasonable to believe that 33 percent is the actual percent of people in the United States whose favorite sport to watch on television is American football

Answer from: Quest

2087568668/1

step-by-step explanation:

Answer from: Quest

$\frac{1}{1815}$

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