Answer :
The 95% confidence interval for the average scores of 16 participants, with an average of 149 and a standard deviation of 46.5, is calculated to be (126.18, 171.83). This interval is determined by the standard error and Z-value for the desired confidence level.
To calculate the 95% confidence interval for the given average scores, we use the formula:
Confidence Interval =[tex]\bar{x} \pm Z \times \frac{s}{\sqrt{n}}[/tex]
Here, [tex]\bar{x}[/tex] is the mean of the scores, s stands for the standard deviation, n represents the number of participants, and Z* is the Z-value corresponding to the desired confidence level, which for 95% confidence is typically 1.96. Plugging the values given:
[tex]\bar{x}[/tex] = 149
s = 46.5
n = 16
Standard Error (SE) =[tex]\frac{s}{\sqrt{n}} = \frac{46.5}{\sqrt{16}} = 11.625[/tex]
Thus, our Confidence Interval is:
[tex]149 \pm 1.96 \times 11.625[/tex]
Calculating the margins gives us:
22.825
Therefore, the 95% confidence interval is:
(149 - 22.825, 149 + 22.825) \= (126.175, 171.825).
This could be that we can say with 95% confidence that the true average score of the population from which this sample is drawn lies within the interval 126.18 and 171.83.