Answer:
a
[tex]\= x = 40.85[/tex]
b
[tex]E = 5.85[/tex]
Ca
[tex]t_c = 2.08 [/tex]
Cb
[tex] t_c = 1.282 [/tex]
Explanation:
From the question we are told that
The sample size is n = 100
The upper limit of the 95% confidence interval is b = 47.2 years
The lower limit of the 95% confidence interval is a = 34.5 years
Generally the sample mean is mathematically represented as
[tex]\= x = \frac{a + b }{2}[/tex]
=> [tex]\= x = \frac{47.2 + 34.5 }{2}[/tex]
=> [tex]\= x = 40.85[/tex]
Generally the margin of error is mathematically represented as
[tex]E = \frac{b- a }{ 2}[/tex]
=> [tex]E = \frac{47.2- 34.5 }{ 2}[/tex]
=> [tex]E = 5.85[/tex]
Considering question C a
From the question we are told the confidence level is 90% , hence the level of significance is
[tex]\alpha = (100 - 90 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
The sample size is n = 22
Given that the sample size is not sufficient enough i.e [tex]n < 30[/tex] we will make use of the student t distribution table
Generally the degree of freedom is mathematically represented as
[tex]df = n- 1[/tex]
=> [tex]df = 22 - 1[/tex]
=> [tex]df = 21[/tex]
Generally from the student t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of 21 is
[tex]t_c =t_{\frac{\alpha }{2} , 21 } = 2.08 [/tex]
Considering question C b
From the question we are told the confidence level is 80% , hence the level of significance is
[tex]\alpha = (100 - 80 ) \%[/tex]
=> [tex]\alpha = 0.20[/tex]
Generally from the normal distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] is
[tex] t_c =Z_{\frac{\alpha }{2} } = 1.282 [/tex]
