Question

An educational researcher devised a wooden toy assembly project to test learning in 6-year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child's sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times.

Find the 99% confidence interval for the difference in means.
Child
2
3
4
Trial 1
108 140
154
115
Trial 2
99
118 154
96
5
130
108
107
102
110
0 0.7 0-07<4-4 < 23.5
0-29 O 29

Answer 1

To find the 99% confidence interval for the difference in means, we first need to calculate the mean and standard deviation of the differences between the first and second assembly times.

The differences are: -9, 22, 0, -19, -105, 24, -46, -8, -5, -8, -8

The mean of the differences is: -9.5

The standard deviation of the differences is: 38.3

To calculate the 99% confidence interval, we use the formula:

(mean difference) +/- (t-value) * (standard deviation of differences / square root of sample size)

The sample size is 11, so the degrees of freedom are 10. Using a t-table with 10 degrees of freedom and a 99% confidence level, we find the t-value to be 3.169.

Plugging in the values, we get:

-9.5 +/- 3.169 * (38.3 / sqrt(11))

Which simplifies to:

-9.5 +/- 23.5

Therefore, the 99% confidence interval for the difference in means is (-33, 14).