Respuesta :
Answer:
The 99% confidence interval would be given by (12.10;12.20)
Step-by-step explanation:
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
[tex]\bar X=12.15[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=0.15[/tex] represent the population standard deviation
n=60 represent the sample size
Assuming the X follows a normal distribution
[tex]X \sim N(\mu, \sigma=0.15[/tex]
The sample mean [tex]\bar X[/tex] is distributed on this way:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
The confidence interval on this case is given by:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex]
Using the normal standard table, excel or a calculator we see that:
[tex]z_{\alpha/2}=2.58[/tex]
Since we have all the values we can replace:
[tex]12.15 - 2.58\frac{0.15}{\sqrt{60}}=12.10[/tex]
[tex]12.15 + 2.58\frac{0.15}{\sqrt{60}}=12.20[/tex]
So on this case the 99% confidence interval would be given by (12.10;12.20)
