Respuesta :
Answer:
The 91% confidence interval estimate of the population mean is between 67.68 and 70.32 cars sold annually.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.91}{2} = 0.045[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.045 = 0.955[/tex], so [tex]z = 1.695[/tex]
Now, find the margin of error M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.695*\frac{16}{\sqrt{420}} = 1.32[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 69 - 1.32 = 67.68.
The upper end of the interval is the sample mean added to M. So it is 69 + 1.32 = 70.32
The 91% confidence interval estimate of the population mean is between 67.68 and 70.32 cars sold annually.
Following are the calculation to the Confidence Interval:
Given:
[tex]\sigma=16\\\\ n=420\\\\ \bar{x}=69[/tex]
To find:
Confidence Interval=?
Solution:
[tex]\sigma=16\\\\ n=420\\\\ \bar{x}=69[/tex]
Here,
[tex]\to C=91\%=0.91[/tex]
[tex]\to\alpha=1-C=1-0.91=0.09[/tex]
Therefore,
[tex]\to Z_c=Z_{\frac{\alpha}{2}}[/tex]
[tex]=Z_{\frac{0.09}{2}} \\\\ =z_{0.045} \\\\ =-1.6954\\\\ = 1.6954 \\\\ =1.7[/tex]
Now, [tex]91\%[/tex] confidence interval:
[tex]C.I. = \bar{x} \pm Z_{c} \times (\frac{\sigma}{\sqrt{(n)}})[/tex]
[tex]= 69 \pm 1.7 \times (\frac{16}{\sqrt{420}} )\\\\= 69 \pm 1.7 \times (\frac{16}{20.49} )\\\\= 69 \pm 1.7 \times ( 0.780)\\\\= 69 \pm 1.326\\\\ =(70.326, 67.674)[/tex]
Therefore, the final answer is "[tex]\bold{(70.326, 67.674)}[/tex]".
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