Question

35. In a simple random sample of 25 high school students, the sample mean of the SAT scores was 1450, and the sample variance was 900. Assume that the data come from a normal distribution , a 95.4 % Confidence interval for the population mean is a. 1450 +/- 180 b. 1450 +/- 18 c. 1450 +/- 12 d. 1450 +/- 360

Answer 1

Answer:

$1450-2.0\frac{30}{\sqrt{25}}=1450-12$    

$1450+2.0\frac{30}{\sqrt{25}}=1450+12$    

And the best option would be:

c. 1450 +/- 12

Step-by-step explanation:

Information provided

$\bar X=1450$ represent the sample mean for the SAT scores

$\mu$ population mean (variable of interest)

$s^2 = 900$ represent the sample variance given

n=25 represent the sample size  

Solution

The confidence interval for the true mean is given by :

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

The sample deviation would be $s=\sqrt{900}= 30$

The degrees of freedom are given by:

$df=n-1=2-25=24$

The Confidence is 0.954 or 95.4%, the value of $\alpha=0.046$ and $\alpha/2 =0.023$, assuming that we can use the normal distribution in order to find the quantile the critical value would be $z_{\alpha/2} \approx 2.0$

The confidence interval would be

$1450-2.0\frac{30}{\sqrt{25}}=1450-12$    

$1450+2.0\frac{30}{\sqrt{25}}=1450+12$    

And the best option would be:

c. 1450 +/- 12