Respuesta :
Answer:
a) The 90% confidence interval for the proportion of teenagers who have 5 or more servings of soft drinks a week is (0.2982, 0.481).
b) 30% = 0.3 is part of the confidence interval, which means that there is no evidence that a higher proportion of teenagers have 5 or more servings of soft drinks a week than the general population.
Step-by-step explanation:
Question a:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
A survey of 77 teenagers finds that 30 have 5 or more servings of soft drinks a week.
This means that [tex]n = 77, \pi = \frac{30}{77} = 0.3896[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3896 - 1.645\sqrt{\frac{0.3896*0.6104}{77}} = 0.2982[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3896 + 1.645\sqrt{\frac{0.3896*0.6104}{77}} = 0.481[/tex]
The 90% confidence interval for the proportion of teenagers who have 5 or more servings of soft drinks a week is (0.2982, 0.481).
Question b:
30% = 0.3 is part of the confidence interval, which means that there is no evidence that a higher proportion of teenagers have 5 or more servings of soft drinks a week than the general population.
