Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The 97% confidence interval is [tex]1528 < \mu < 1612 [/tex]
b
Option A and Option D are correct
Step-by-step explanation:
From the question we are told that
The sample size is n = 119
The sample mean is [tex]\$ 1570[/tex]
The population standard deviation is [tex]\sigma = \$ 211[/tex]
Generally given that the confidence level is 97% then the level of significance is mathematically represented as
[tex]\alpha = (100 - 97 )\%[/tex]
=> [tex]\alpha = 0.03[/tex]
From the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 2.17[/tex]
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }[/tex]
[tex]E = 2.17* \frac{211}{\sqrt{ 119 } }[/tex]
=> [tex]E = 41.98 [/tex]
Generally the 97% confidence interval is mathematically represented as
[tex]\= x - E < \mu < \= x + E[/tex]
=> [tex]1570 - 41.98 < \mu < 1570 + 41.98 [/tex]
=> [tex]1528 < \mu < 1612 [/tex]
Generally looking at confidence interval we see that the width is dependent on the size of the margin of error.
Generally the margin error is directly proportional to the critical value of [tex]\frac{\alpha }{2}[/tex] which increases when the confidence level increases and vise versa
Also the margin error is inversely proportional to the square root of sample size
Hence to reduce the width of the confidence interval we need to lower the confidence level and increase the sample size
