Question

A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for the population mean μ is between_________.

Answer 1

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 ($\mu$ ):-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size.

s= Sample standard deviation

$\overline{x}$ = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

$\overline{x}=84$

n= 64

Degree of freedom : df = n-1=63  

s= 5.5

Significance level : $\alpha=1-0.90=0.1$

Two-tailed T-value for df = 63 and  $\alpha=1-0.90=0.1$ would be

$t_{\alpha/2,df}=t_{0.05,63}=1.669$  (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

$84\pm (1.669)\dfrac{5.5}{\sqrt{64}}$

$=84\pm (1.669)\dfrac{5.5}{8}$

$\approx84\pm 1.15$

$=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)$

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,