Question

In a random sample of 64 trout, the average weight is 3 pounds, with standard deviation s = .5 pounds Find:

sample mean

standard error of the sample mean

margin of error 95%

95% confidence interval for the population mean, µ:

Answer 1

Solution :

Given that,

$\bar x$ = 3

s = 0.5

n = 64

The standard error = (s /$\sqrt$n)

= (0.5 / $\sqrt$ 64)

=0.06

The standard error =0.06

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 95% confidence level the t is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,63 =1.998

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 1.998 * (0.5 / $\sqrt$ 64)

= 0.12

Margin of error =0.12

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

3 - 0.12 < $\mu$ < 3 + 0.12

2.88 < $\mu$ < 3.12

(2.88, 3.12 )