Question

Hardy wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. He randomly selects a sample of 98 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 8.2 and a standard deviation of 3.3. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies?

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.2

Population standard deviation =    = 3.3
Sample size = n =98

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2    * ( /n)

= 1.645 * (3.3 /  98 )

= 0.55
At 90% confidence interval
is,

- E < < + E

8.2- 0.55 <   < 8.2 + 0.55

4.65 <   < 8.75