Question

You measure 47 turtles' weights, and find they have a mean weight of 42 ounces. Assume the sample weights' standard deviation is 6.2 ounces. Based on this, construct a 90% confidence interval for the true population mean turtle weight.

Give your answers as decimals, to 2 decimal places

Answer 1

b )solution

Given that,

= 42

s =6.2

n = 47

Degrees of freedom = df = n - 1 = 47 - 1 = 46

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t 0.05,46 =  1.679 ( using student t table)

Margin of error = E = t/2,df * (s /n)

=  1.679 * ( 6.2/ 47) = 1.52

The 90% confidence interval is,

- E < < + E

42 - 1.52 < < 42 + 1.52

40.48< <43.52