Question

Imagine 126 people complete a test. An analysis of scores for the test reveals M = 67.00, SD = 12.00, and r x x = 0.73. Emily obtained a score of 83 on the test. Calculate a 95 percent confidence interval for Emily's true score. For this question, center the CI on Xest (as per the textbook approach) not X observed.

Answer 1

The 95 percent confidence interval for Emily's true score on the test is approximately (78.17, 87.83).

Step-by-step explanation:

The confidence interval for Emily's true score can be calculated using the formula:

`\[ \text{CI} = \bar{X}_{\text{est}} \pm t \left((s)/(√(N))\right) \]`

where:

`\( \bar{X}_{\text{est}} \)` is the mean of the sample scores,

t is the critical t-value based on the degrees of freedom
`(\(df = N - 1\))` and desired confidence level,

s is the standard deviation of the sample scores,

N is the sample size.

Given:

`\[ \bar{X}_{\text{est}} = 67.00 \]`

s = 12.00

N = 126

`\[ r_(xx) = 0.73 \]`

First, we need to find the standard error of the mean SEM :

`\[ SEM = (s)/(√(N)) \]`

`\[ SEM = (12.00)/(√(126)) \approx 1.071 \]`

Next, we find the critical t-value for a 95% confidence interval with
`\( df = N - 1 = 125 \)`. Consulting a t-table or using statistical software, we find
`\( t \approx 1.980 \)`.

Now, substitute these values into the formula:

`\[ \text{CI} = 67.00 \pm 1.980 * 1.071 \]`

`\[ \text{CI} \approx (78.17, 87.83) \]`

Therefore, we can be 95 percent confident that Emily's true score on the test falls within the interval (78.17, 87.83).