Answer: The 90% confidence interval for the population mean μ is between 82.85 and 85.15,
Step-by-step explanation:
When population standard deviation is not given ,The confidence interval population proportion is given by ([tex]\mu[/tex] ):-
[tex]\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}[/tex]
, where n= Sample size.
s= Sample standard deviation
[tex]\overline{x}[/tex] = sample mean
t* = Critical t-value (Two-tailed)
As per given , we have
[tex]\overline{x}=84[/tex]
n= 64
Degree of freedom : df = n-1=63
s= 5.5
Significance level : [tex]\alpha=1-0.90=0.1[/tex]
Two-tailed T-value for df = 63 and [tex]\alpha=1-0.90=0.1[/tex] would be
[tex]t_{\alpha/2,df}=t_{0.05,63}=1.669[/tex] (By t-distribution table)
i.e. t*= 1.669
The 90% confidence interval for the population mean μ would be
[tex]84\pm (1.669)\dfrac{5.5}{\sqrt{64}}[/tex]
[tex]=84\pm (1.669)\dfrac{5.5}{8}[/tex]
[tex]\approx84\pm 1.15[/tex]
[tex]=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)[/tex]
∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,
