Question

Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 79 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 19.7 and a standard deviation of 1.9. What is the 95% 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

(19.3, 20.1)

Explanation:

• Sample Mean = 19.7

,

• Sample Standard Deviation = 1.9

,

• Sample Size = 79

Since the population standard deviation is not known, we estimate it from the sample standard deviation and then use a t-students distribution to calculate the critical value.

The formula for estimating the population standard deviation is:

`\sigma=(s)/(√(N))=(1.9)/(√(79))=0.21`

The degrees of freedom for this sample size is:

`df=N-1=79-1=78`

The t-value for a 95% confidence interval and 78 degrees of freedom is t=1.99.

Thus, the margin of error is:

`MOE=t*\sigma=1.99*0.21=0.42`

Thus, the lower and upper bounds of the confidence interval are:

`\begin{gathered} Lower:Mean-MOE=19.7-0.42=19.3 \\ Upper:Mean+MOE=19.7+0.42=20.1 \end{gathered}`

The confidence interval is (19.3, 20.1).