Respuesta :
Answer:
The margin of error for 90% confidence interval is 2.98, for 95% confidence interval is 3.55 and for 99% confidence interval is 4.68.
Step-by-step explanation:
The sample size is, n = 400.
The maximum value is, Max. = $225
The minimum value is, Min. = $80
The standard deviation of a distribution using the range of the data is:
[tex]SD=\frac{Range}{4}=\frac{Max.-Min.}{4}=\frac{225-80}{4}= 36.25[/tex]
As the sample is large, i.e. n = 400 > 30, according to the central limit theorem the sampling distribution of sample mean will follow a normal distribution.
The formula to compute the margin of error is:
[tex]MOE=z_{\alpha /2}\times \frac{SD}{\sqrt{n}}[/tex]
- For a 90% confidence interval:
The value of α is 1 - 0.90 = 0.10
The critical value is,
[tex]z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645[/tex]
(Use the standard normal table)
The MOE is:
[tex]MOE=z_{\alpha /2}\times \frac{SD}{\sqrt{n}}\\=1.645\times \frac{36.25}{\sqrt{400}} \\=2.98[/tex]
- For a 95% confidence interval:
The value of α is 1 - 0.95 = 0.05
The critical value is,
[tex]z_{\alpha /2}=z_{0.05/2}=z_{0.025}=1.96[/tex]
(Use the standard normal table)
The MOE is:
[tex]MOE=z_{\alpha /2}\times \frac{SD}{\sqrt{n}}\\=1.96\times \frac{36.25}{\sqrt{400}} \\=3.55[/tex]
- For a 99% confidence interval:
The value of α is 1 - 0.99 = 0.01
The critical value is,
[tex]z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58[/tex]
(Use the standard normal table)
The MOE is:
[tex]MOE=z_{\alpha /2}\times \frac{SD}{\sqrt{n}}\\=2.58\times \frac{36.25}{\sqrt{400}} \\=4.68[/tex]
Thus, the margin of error for 90% confidence interval is 2.98, for 95% confidence interval is 3.55 and for 99% confidence interval is 4.68.
