Respuesta :
Using the z-distribution, we have that:
a) The estimate is of 0.294.
b) Since the entire confidence interval is below 0.5, there is enough evidence that the minority of the population would say yes.
c) The 99% confidence interval for the population proportion who would say “yes" is (0.26, 0.328), which means that we are 99% sure that the true population proportion is between these two values.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
- z is the critical value.
- n is the sample size.
In this problem, we have a 99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so the critical value is z = 2.575.
The estimate and the sample size are given, respectively, by:
[tex]\pi = \frac{344}{1170} = 0.29, n = 1170[/tex]
The bounds of the confidence interval are:
- [tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.294 - 2.575\sqrt{\frac{0.294(0.706)}{1170}} = 0.26[/tex]
- [tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.294 + 2.575\sqrt{\frac{0.294(0.706)}{1170}} = 0.328[/tex]
The 99% confidence interval for the population proportion who would say “yes" is (0.26, 0.328), which means that we are 99% sure that the true population proportion is between these two values.
Since the entire confidence interval is below 0.5, there is enough evidence that the minority of the population would say yes.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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