High School

The scores on the Wechsler Intelligence Scale for Children (WISC) are known to be normally distributed with a standard deviation of 10. A simple random sample of 45 children from this population is taken, and each child is given the WISC test. The mean of the 45 scores is 99.8.

Find a 99% confidence interval for the mean score.

Using your answer, write the confidence interval in a complete sentence related to the scenario.

Answer :

The 99% confidence interval for the mean of the scores on the WISC is approximately 95.96 to 103.64.

To find a 99% confidence interval for the mean of the scores on the Wechsler Intelligence Scale for Children (WISC), we can use the formula:

Confidence interval = sample mean ± (critical value) * (standard deviation / √sample size)

First, we need to find the critical value associated with a 99% confidence level. Since the sample size is large (n > 30), we can use the Z-table to find the critical value. For a 99% confidence level, the critical value is approximately 2.576.

Next, we plug in the values into the formula:

Confidence interval = 99.8 ± (2.576) * (10 / √45)

Calculating the values:

Confidence interval = 99.8 ± (2.576) * (10 / √45)

Confidence interval = 99.8 ± (2.576) * (10 / 6.708)

Confidence interval = 99.8 ± (2.576) * 1.491

Now, we can calculate the lower and upper bounds of the confidence interval:

Lower bound = 99.8 - (2.576) * 1.491

Upper bound = 99.8 + (2.576) * 1.491

Calculating the values:

Lower bound ≈ 99.8 - 3.84

Lower bound ≈ 95.96

Upper bound ≈ 99.8 + 3.84

Upper bound ≈ 103.64

Therefore, the 99% confidence interval for the mean of the scores on the WISC is approximately 95.96 to 103.64.

In the context of this scenario, we can say with 99% confidence that the true mean score on the WISC for the population of children lies between 95.96 and 103.64.

Learn more about critical value from the below link

https://brainly.com/question/32591251

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