Respuesta :
Answer:
a) The sample proportion is of 0.6.
b) The 95% confidence interval for the proportion of potential customers who prefer your product is (0.4642, 0.7358).
c) A sample size of 145 is needed.
Step-by-step explanation:
(a) Find the sample proportion.
30 out of 50. So
[tex]\pi = \frac{30}{50} = 0.6[/tex]
The sample proportion is of 0.6.
(b) Find the 95% confidence interval for the proportion of potential customers who prefer your product.
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].
For this problem, we have that:
[tex]n = 50, \pi = 0.6[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6*0.4}{50}} = 0.4642[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6*0.4}{50}} = 0.7358[/tex]
The 95% confidence interval for the proportion of potential customers who prefer your product is (0.4642, 0.7358).
(c) If you want the 95% maximum likely error to be 0.08 or less, what would you choose for a sample size
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
We need a sample size of n, and n is found when M = 0.08. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.08 = 1.96\sqrt{\frac{0.6*0.4}{n}}[/tex]
[tex]0.08\sqrt{n} = 1.96*sqrt{0.6*0.4}[/tex]
[tex]\sqrt{n} = \frac{1.96*sqrt{0.6*0.4}}{0.08}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*sqrt{0.6*0.4}}{0.08})^2[/tex]
[tex]n = 144.1[/tex]
Rounding up
A sample size of 145 is needed.
