Question

A sample of 10 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was 3 ounces with a sample standard deviation s=0.5 ounces. We would like to calculate an 80% confidence interval for the average weight of a sample of size 10 .

(3%) standard error =   

(3%) The critical t value for an 80% confidence interval is

tcrit=   

(3%) EBM=   

(3%) An 80% confidence interval for the population average weight of the candies is from

to   

Answer 1

Solution :

Given that,

= 3

s =0.5

n =10

Degrees of freedom = df = n - 1 =10 - 1 = 9

a )

standard error =(s /n) = (0.5 / 10) =0.36

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

critical value

t /2,df = t0.025,9 = 2.262 ( using student t table)

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

= 2.262 * (0.5 / 10) =0.36

The 80% confidence interval is,

- E < < + E

3 -0.36 < <3 + 0.36

2.64 < < 3.36

(2.64 , 3.36 )