Question

R8.5. Smart kids A school counselor wants to know how smart the students in her school are. She gets funding from the principal to give an IQ test to an SRS of $60$of the over $1000$students in the school. The mean IQ score was $114.98$and the standard deviation was $14.80$.
(a) Describe the parameter $\mu$ in this setting.
(b) Explain why the Normal condition is met in this case.
(c) Construct a $90\%$ confidence interval for the mean IQ score of students at the school.
(d) Interpret your result from part (c) in context.

Short answer

(a) The parameter $\mu$displays the population mean, which displays the average IQ score of all students enrolled in school.

(b) Because the sample size of $60$is greater than $30$, the Normal condition is satisfied, and the sample distribution can be presumed to be approximately normal.
(c) The confidence interval is $(111.79,118.17)$.

(d) There is a $90\%$confident that the genuine mean IQ score of the students will fall somewhere between $111.79$and $118.17$.

Step-by-step solution

Part (a) Step 1: Given information

To describe the parameter $\mu$in this setting.

Part (a) Step 2: Explanation

Let, the sample mean is $\left(\overline{x}\right)=114.98$.
The sample standard deviation is $\left(s\right)=14.80$.
And the Sample size is $\left(n\right)=60$.
The confidence interval is calculated as:
$\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}<\mu <\overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}$
The parameter $\mu$displays the population mean, which displays the average IQ score of all students enrolled in school.

Part (b) Step 1: Given information

To explain that why the Normal condition is met in this case.

Part (b) Step 2: Explanation

The sample size must be more than $30$ to satisfy the normalcy criteria.
The sample size in this case is $60$, which is higher than $30$.
As a result, the criteria of normality is met.

Part (c) Step 1: Given information

To construct a $90\%$confidence interval for the mean IQ score of students at the school.

Part (c) Step 2: Explanation

Using a Ti-83 calculator, the following confidence interval was calculated:

As a result, the confidence interval is $(111.79,118.17)$.

Part (d) Step 1: Given information

To interpret your result from part (c) in context.

Part (a) Step 2: Explanation

As the result from part (c) as:

There is a $90\%$confident that the genuine mean IQ score of the students will fall somewhere between $111.79$and $118.17$.