Question

The Weschler Intelligence Scale for Children (WISC) is an intelligence test designed for children between the ages of 6 and 16. The test is standardized so that the mean score for all children is 100 and the standard deviation is 15.
Suppose that the administrators of a very large and competitive school district wish to estimate the mean WISC score for all students enrolled in their programs for gifted and talented children. They obtained a random sample of 40 students currently enrolled in at least one program for gifted and talented children. The test scores for this sample are as follows:
126,111,141,96,123,115,116,113,127,114,122,116,104,126,126,132,139,111,143,121,119,107,154,125,120,135,142,132,94,103,143,115,132,107,118,105,101,135,94,153
a. Use this data to calculate the mean WISC score, x⎯⎯, for these 40 students.
b. Next, compute the standard error of the mean, SE, assuming that the standard deviation of WISC scores for students in the district is the same as for the population as a whole.
c. Finally, determine both the lower and upper bounds of a 90% confidence interval for μ, the mean score for all students in the school district who are enrolled in gifted and talented programs.

Answer 1

Answer:

a) $\bar X = 121.4$

b) $SE = \frac{\sigma}{\sqrt{n}}= \frac{15}{\sqrt{40}}=2.372$

c) $121.4-1.64\frac{15}{\sqrt{40}}=117.51$    

$121.4+1.64\frac{15}{\sqrt{40}}=125.29$  

Step-by-step explanation:

Data given: 126,111,141,96,123,115,116,113,127,114,122,116,104,126,126,132,139,111,143,121,119,107,154,125,120,135,142,132,94,103,143,115,132,107,118,105,101,135,94,153

Part a

For this case we can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And after replace we got:

$\bar X = 121.4$

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

For this case we assume that the population deviation is $\sigma=15$, we can calculate the standard error with this formula:

$SE = \frac{\sigma}{\sqrt{n}}= \frac{15}{\sqrt{40}}=2.372$

Part c

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$121.4-1.64\frac{15}{\sqrt{40}}=117.51$    

$121.4+1.64\frac{15}{\sqrt{40}}=125.29$  

Answer 2

Answer:

attached below

Step-by-step explanation: