Question

Sound it out: Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 133 first-graders who were learning English were asked to identify as many letter sounds as possible in a period of one minute. The average number of letter sounds identified was 34.21 with a standard deviation of 23.33.

Construct a 99.9% confidence interval for the mean number of letter sounds identified in one minute. Round the answers to two decimal places

Answer 1

The 99.9% confidence interval for the mean number of letter sounds identified in one minute if between 0 and 113.07.

Explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 133 - 1 = 132

99.9% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 132 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.999)/(2) = 0.9995`. So we have T = 3.38

The margin of error is:

M = T*s = 3.38*23.33 = 78.86

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 34.21 - 78.86. They can't identify a negative number of letter sounds, so the lower end is 0.

The upper end of the interval is the sample mean added to M. So it is 34.21 + 78.86 = 113.07

The 99.9% confidence interval for the mean number of letter sounds identified in one minute if between 0 and 113.07.