The scores on the Wechsler Intelligence Scale for Children (WISC)are thought to be Normally distributed with a standard deviation of $\sigma$= 10.A simple random sample of 25 children is taken,and each is given the WISC.The mean of the 25 scores is = 104.32.Based on these data,what is a 95% confidence interval for $\mu$?
A)104.32 ± 0.78
B)104.32 ± 3.29
C)104.32 ± 3.92
D)104.32 ± 19.60

Question

The scores on the Wechsler Intelligence Scale for Children (WISC)are thought to be Normally distributed with a standard deviation of $\sigma$= 10.A simple random sample of 25 children is taken,and each is given the WISC.The mean of the 25 scores is  = 104.32.Based on these data,what is a 95% confidence interval for $\mu$?
A)104.32 ± 0.78 
B)104.32 ± 3.29 
C)104.32 ± 3.92 
D)104.32 ± 19.60

Quizplus · Accepted Answer

We can use the formula for a confidence interval for the population mean:

Confidence interval = sample mean ± (t-value)*(standard error)

where the t-value is based on the degrees of freedom (df = n-1) and the desired level of confidence. For a 95% confidence interval with df = 24, the t-value is 2.064.

The standard error is calculated as standard deviation / sqrt(sample size), which gives:

standard error = 10 / sqrt(25) = 2

Plugging in the values, we get:

Confidence interval = 104.32 ± (2.064)*(2) = 104.32 ± 4.13

So the 95% confidence interval is 104.32 ± 3.92 (rounded to two decimal places), which corresponds to choice C.

The Scores on the Wechsler Intelligence Scale for Children (WISC)are σ\sigmaσ

The Scores on the Wechsler Intelligence Scale for Children (WISC)are $\sigma$