Respuesta :
Answer:
[tex]12.1-1.64\frac{0.35}{\sqrt{50}}=12.019[/tex]
[tex]12.1+1.64\frac{0.35}{\sqrt{50}}=12.181[/tex]
So on this case the 90% confidence interval would be given by (12.019;12.181)
And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied
Step-by-step explanation:
Previous concepts
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".
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".
[tex]\bar X=12.1[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=0.35[/tex] represent the population standard deviation
n=50 represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]
Now we have everything in order to replace into formula (1):
[tex]12.1-1.64\frac{0.35}{\sqrt{50}}=12.019[/tex]
[tex]12.1+1.64\frac{0.35}{\sqrt{50}}=12.181[/tex]
So on this case the 90% confidence interval would be given by (12.019;12.181)
And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied
