Answer:
[tex]1450-2.0\frac{30}{\sqrt{25}}=1450-12[/tex]
[tex]1450+2.0\frac{30}{\sqrt{25}}=1450+12[/tex]
And the best option would be:
c. 1450 +/- 12
Step-by-step explanation:
Information provided
[tex]\bar X=1450[/tex] represent the sample mean for the SAT scores
[tex]\mu[/tex] population mean (variable of interest)
[tex]s^2 = 900[/tex] represent the sample variance given
n=25 represent the sample size
Solution
The confidence interval for the true mean is given by :
[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)
The sample deviation would be [tex]s=\sqrt{900}= 30[/tex]
The degrees of freedom are given by:
[tex]df=n-1=2-25=24[/tex]
The Confidence is 0.954 or 95.4%, the value of [tex]\alpha=0.046[/tex] and [tex]\alpha/2 =0.023[/tex], assuming that we can use the normal distribution in order to find the quantile the critical value would be [tex]z_{\alpha/2} \approx 2.0[/tex]
The confidence interval would be
[tex]1450-2.0\frac{30}{\sqrt{25}}=1450-12[/tex]
[tex]1450+2.0\frac{30}{\sqrt{25}}=1450+12[/tex]
And the best option would be:
c. 1450 +/- 12
