Use the reference below when solving the questions.

Estimating a Population Mean (Average)

In the previous lesson we calculated the margin of error for proportions (percentages). Now we turn to calculating the margin of error for means (averages).

In order to continue we must learn to read a new table,

entitled thet-Distribution: Critical t Values.

Just as calculating the margin of error for percentages required a Z/2 based on a confidence level, calculating the margin of error for the mean requires at/2 based on a confidence level.

Thet/2 values that we will need are found by taking two values to the t-table:

1) degrees of freedom

2) error (denoted , pronounced alpha)

Degrees of freedom are equal to sample size, n, minus 1.

For example: If sample size is 50, there are 49 degrees of freedom.

If the specified confidence level (cl) is 95%, then the error () is the other 5%. The error, , is expressed in decimal form.

Aclof 90% leads to = 0.10

Aclof 95% leads to = 0.05

Aclof 98% leads to = 0.02

Example 1: A sample size, n, of 10 and a confidence level of 95%

degrees of freedom = 9

= 0.05 (remember, is the total area in two tails)

Find 9 in the left most, degrees of freedom, column.

Find 0.05 in the Area in Two Tails row.

At the intersection you find 2.262. That is thet/2 for n = 10 andcl= 95%

Referring to the reference above, solve the following questions and show your work.

1) We have a sample of 40 chocolate chip cookies. The sample mean of chips per cookie is 23.95. The sample standard deviation is 2.55 chips. Construct a 99% confidence interval estimate of the population mean of chocolate chips in such cookies.

2) We have a sample of 195 newborn boys. The sample mean for weight is 32.7 hectograms and the sample standard deviation is 6.6 hectograms. Construct a 95% confidence interval estimate of the mean birth weight for the population of baby boys.

3) We have 7 samples of sushi from different stores in New York City. The sample mean for mercury, in parts-per-million, is 0.719. The sample standard deviation is 0.366 parts-per-million. Construct a 90% confidence interval estimate of the average amount of mercury, in parts-per-million, for the population of sushi in New York stores.