Answer :
Answer:
The 90% confidence interval for the proportion of college students who drive sticks is (0.173, 0.293).
Step-by-step explanation:
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 zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
A random sample of college students who owned cars revealed the following: out of 133 cars, 31 had stick shifts.
This means that [tex]n = 133, \pi = \frac{31}{133} = 0.233[/tex]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue 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}} = 0.233 - 1.645\sqrt{\frac{0.233*0.767}{133}} = 0.173[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.233 + 1.645\sqrt{\frac{0.233*0.767}{133}} = 0.293[/tex]
The 90% confidence interval for the proportion of college students who drive sticks is (0.173, 0.293).