Respuesta :
Using the z-distribution, the 95% confidence interval for the proportion of unsatisfactory meals is given by:
(0.0553, 0.0647).
What is a confidence interval of proportions?
A confidence interval of proportions is given by the following rule:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion, which is also the estimate of the parameter.
- z is the critical value.
- n is the sample size.
The confidence level is of 95%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.
The estimate and the sample size of the interval are given as follows:
[tex]\pi = \frac{600}{10000} = 0.06, n = 10000[/tex]
The lower bound of the interval is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.06 - 1.96\sqrt{\frac{0.06(0.94)}{10000}} = 0.0553[/tex]
The upper bound of the interval is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.06 + 1.96\sqrt{\frac{0.06(0.94)}{10000}} = 0.0647[/tex]
Hence the interval in (LCL, UCL) format is:
(0.0553, 0.0647).
What is the missing information?
The problems asks for the 95% confidence interval of unsatisfactory meals, considering that, out of 10000 meals, 600 were unsatisfactory.
More can be learned about the z-distribution at https://brainly.com/question/25890103
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