Question

A random sample of 64 observations produced a mean value of 86 and standard deviation of 4.5. The 95% confidence interval for the population mean μ is between:_________.

Answer 1

Answer: (84.876, 87.124)

Step-by-step explanation:

Confidence interval for population mean if population standard deviation is unknown:

$\overline{x}\pm t_{\alpha/2}(\dfrac{s}{\sqrt{n}})$

, where n= sample size

s= sample standard deviation

$\overline{x}$ = sample mean

$\alpha=$ significance level

$t_{\alpha/2}$ = critical-t value

Given:  n= 64

Degree of freedom = n-1 = 63

s= 4.5

$\overline{x}$ = 86

$\alpha=$ 0.05

$t_{\alpha/2}$ = 1.9983

Now, the required 95% confidence interval would be:

$86\pm (1.9983)(\dfrac{4.5}{\sqrt{64}})\\\\=86\pm (1.9983)(\dfrac{4.5}{8})\\\\=86\pm (1.9983)(0.5625)\\\\\approx86\pm 1.1240\\\\ =(86-1.1240,\ 86+1.1240)\\\\=(84.876,\ 87.124)$

The 95% confidence interval for the population mean μ is between: (84.876, 87.124)