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

`1450-2.0(30)/(√(25))=1450-12`

`1450+2.0(30)/(√(25))=1450+12`

And the best option would be:

c. 1450 +/- 12

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)(\sigma)/(√(n))` (1)

The sample deviation would be
`s=√(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(30)/(√(25))=1450-12`

`1450+2.0(30)/(√(25))=1450+12`

And the best option would be:

c. 1450 +/- 12