Answer:
The 90% confidence interval is [tex] 87.95 < \mu < 92.05 [/tex]
Step-by-step explanation:
From the question we are told that
The sample size is [tex]n = 28[/tex]
The sample mean is [tex]\= x = 90[/tex]
The standard deviation is [tex]s= 60[/tex]
Generally given that the sample size is not large enough i.e n < 28 then we will make use of the t distribution table
Generally the degree of freedom is mathematically represented as
[tex]df = n- 1[/tex]
=> [tex]df = 28 - 1[/tex]
=> [tex]df = 2 7[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally from the student t distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 2 7[/tex] is
[tex]t_{\frac{\alpha }{2} , 27 } = 2.052[/tex]
Generally 95% confidence interval is mathematically represented as
[tex]\= x -E < \mu < \=x +E[/tex]
=> [tex]90 - 2.052 < \mu < 90 + 2.052 [/tex]
=> [tex] 87.95 < \mu < 92.05 [/tex]
