Answer:
(a) The 95% confidence interval representing the percentage of the entire U.S. population that would select American football as their preferred television sport is (0.34, 0.40).
(b) Not reasonable.
Step-by-step explanation:
Given:
n = 1000
= 0.37
(a)
The confidence interval (1 - α)% for proportion p is:
Where:
= the sample proportion
n = sample size
= critical z value.
Calculate the critical value of z for a 95% confidence level as shown:
*Refer to a z-table for the necessary value.
Compute the 95% confidence interval for proportion p as follows:
Thus, the 95% confidence interval indicating the proportion of individuals in the U.S. who might say their preferred sport on TV is American football is (0.34, 0.40).
(b)
Next, we must assess if it's rational to consider that the actual percent of people in the United States who favor American football on television is 33%.
The hypothesis is defined as:
H₀: The portion of the U.S. population claiming American football as their favorite sport on television is 33%, meaning p = 0.33.
Hₐ: The proportion of people in the U.S. whose favorite sport to watch on television is not 33%, or p ≠ 0.33
This hypothesis can be verified using a confidence interval.
The decision rule:
If the (1 - α)% confidence interval contains the null value of the hypothesis, then the null hypothesis is not rejected. If, however, the (1 - α)% confidence interval excludes the null value of the hypothesis, then the null hypothesis is rejected.
<pthe confidence="" interval="" for="" the="" proportion="" of="" all="" u.s.="" individuals="" indicating="" that="" american="" football="" is="" their="" favorite="" sport="" on="" television="">
The confidence interval does encompass the null value of p, which is 0.33.
<pthus the="" null="" hypothesis="" will="" be="" rejected.=""><pin conclusion="" it="" is="">not reasonable to accept that 33% represents the actual percentage of those in the U.S. whose favorite televised sport is American football.
</pin></pthus></pthe>
