Answer:
15.33[tex]\leq[/tex]μ[tex]\leq[/tex]15.89
Step-by-step explanation:
given observations
15.4
15.8
15.4
15.1
15.8
15.9
15.8
15.7.
mean(x)=15.6125
Variance (s2): 0.068593750000048
standard deviation(s)=0.2619040854970537
as we dont know the population standard deviation so we use t-stat
formula
t[tex]\frac{α }{2}[/tex],n-1=[tex]\frac{x-μ}{[tex]\frac{s}{[tex]\sqrt{n}[/tex]}[/tex]}[/tex]
where
s-sample standard deviation
x-sample mean
μ-population mean
n-sample size
x-t×[tex]\frac{s}{[tex]\sqrt{n}[/tex]}[/tex][tex]\leq[/tex]μ[tex]\leq[/tex] t×[tex]\frac{s}{[tex]\sqrt{n}[/tex]}[/tex]+x
for 98% confidence interval and 7 degrees of freedom
t=2.998
15.33[tex]\leq[/tex]μ[tex]\leq[/tex]15.89 is the 98% confidence interval for mean
