Answer:
a) [tex]34\% \:<\:p\:<\:40\%[/tex]
b) No, it is not reasonable
Step-by-step explanation:
According to the survey, 37 percent indicated that their favorite sport to watch on television was American football.
This means that: [tex]\hat p=\frac{37}{100}=0.37[/tex]
and [tex]\hat q=1-0.37=0.36[/tex]
Since [tex]\alpha=1-0.95=0.05[/tex], we have: [tex]z_{\frac{\alpha}{2} }=1.96[/tex]
Substitute in the formula:
[tex]\hat p- z_{\frac{\alpha}{2} }\sqrt{\frac{\hat p \hat q}{n} } \:<\:p\:<\:\hat p+ z_{\frac{\alpha}{2} }\sqrt{\frac{\hat p \hat q}{n} }[/tex]
This gives us:
[tex]0.37- 1.96\times \sqrt{\frac{(0.37)(0.63)}{1000} } \:<\:p\:<\:0.37+ 1.96\times \sqrt{\frac{(0.37)(0.63)}{1000} }[/tex]
We evaluate to get:
[tex]0.37- 1.96\times 0.0153 \:<\:p\:<\:0.37+ 1.96\times 0.0153[/tex]
We simplify to get:
[tex]0.37- 0.029988 \:<\:p\:<\:0.37+ 0.029988[/tex]
This gives:
[tex]0.340012 \:<\:p\:<\:0.399988[/tex]
Or
[tex]34\% \:<\:p\:<\:40\%[/tex]
Part b)
From a) we obtain the confidence interval as [tex]34\% \:<\:p\:<\:40\%[/tex].
This means we are 95% confident that, the actual percent of people in the United States whose favorite sport to watch on television is American football is between 34% and 40%.
Since 33% is less than 34%, 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
