Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]33.55 < \mu < 35.5[/tex]
b
[tex]34.03 < \mu < 34.969 [/tex]
c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
d
The correct option is E
Step-by-step explanation:
From the question we are told that
The sample mean is [tex]\= x = 34.5[/tex]
The standard deviation is [tex]s = 3.4[/tex]
Generally given that the confidence level is 95% then the level of significance is
[tex]\alpha = (100 - 95)\%[/tex]
=> [tex]\alpha = 0.05 [/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex]Z_{\frac{\alpha }{2} } = 1.96[/tex]
Considering question a
From the question n = 49
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{49} }[/tex]
=> [tex]E = 0.952 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.952 < p < 34.5 + 0.952[/tex]
=> [tex]33.55 < \mu < 35.5[/tex]
Considering question b
From the question n = 196
Generally the margin of error is mathematically represented as
[tex]E = Z_{\frac{\alpha }{2} } * \frac{s }{\sqrt{n} }[/tex]
=> [tex]E = 1.96* \frac{ 3.4 }{\sqrt{196} }[/tex]
=> [tex]E = 0.4687 [/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < p < \=x +E[/tex]
[tex]34.5 -0.4687 < p < 34.5 +0.4687[/tex]
=> [tex]34.03 < \mu < 34.969 [/tex]
Considering question c
Generally the width at n = 49 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.952 [/tex]
[tex]w = 1.904 [/tex]
Generally the width at n = 196 is mathematically represented as
[tex]w = 2 * E[/tex]
[tex]w = 2 * 0.4687 [/tex]
[tex]w = 0.9374 [/tex]
Now when the sample size is quadrupled i.e from n = 49 to n = 196
The width of the confidence interval decrease by 2 from 1.904 to 0.9374
