Question

If n=14, ¯xx¯(x-bar)=45, and s=4, construct a confidence interval at a 99% confidence level. Assume the data came from a normally distributed population.

Give your answers to one decimal place.

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 65 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 7.1 and a standard deviation of 3.9. What is the 90% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 45

sample standard deviation = s = 4

sample size = n = 14

Degrees of freedom = df = n - 1 = 14 - 1 = 13

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = 3.012

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

= 3.012 * ( 4 / 14)

Margin of error = E = 3.2

The 99% confidence interval estimate of the population mean is,

  ± E

45 ± 3.2

941.8 , 48.2)

Given that,

Point estimate = sample mean = = 7.1

sample standard deviation = s = 3.9

sample size = n = 65

Degrees of freedom = df = n - 1 = 65 - 1 = 64

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

t/2,df = 1.669

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

= 1.669 * (3.9 / 65)

Margin of error = E = 0.8

The 90% confidence interval estimate of the population mean is,

  ± E  

7.1 ± 0.8

(6.3 , 7.9)