Respuesta :
Answer:
The 95 % confidence interval for the population mean test score for high school seniors is between 1101.8 and 1118.2. This means that we are 95% sure that the true population mean test score for high school seniors is between 1101.8 and 1118.2.
Step-by-step explanation:
Large sample size, so we can build a normal confidence interval.
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.96*\frac{132}{\sqrt{1000}} = 8.2[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 1110 - 8.2 = 1101.8.
The upper end of the interval is the sample mean added to M. So it is 1110 + 8.2 = 1118.2.
The 95 % confidence interval for the population mean test score for high school seniors is between 1101.8 and 1118.2. This means that we are 95% sure that the true population mean test score for high school seniors is between 1101.8 and 1118.2.
Answer:
95% confidence interval for the population mean test score for high school seniors is (1101.82 , 1118.18).
Step-by-step explanation:
We are given that a new version of the SAT is given to 1000 randomly selected high school seniors. The sample mean test score is 1110, and the sample standard deviation is 132.
Assuming data follows normal distribution.
So, firstly the pivotal quantity for 95% confidence interval for the population mean test score is given by;
P.Q. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean test score = 1110
[tex]\mu[/tex] = population mean test score
s = sample standard deviation = 132
n = sample size = 1000
So, 95% confidence interval for population mean, [tex]\mu[/tex] is;
P(-1.96 < [tex]t_9_9_9[/tex] < 1.96) = 0.95
P(-1.96 < [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P( [tex]-1.96 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]{\bar X-\mu}[/tex] < [tex]1.96 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
P( [tex]\bar X-1.96 \times {\frac{s}{\sqrt{n} } }[/tex] < [tex]\mu[/tex] < [tex]\bar X +1.96 \times {\frac{s}{\sqrt{n} } }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = ( [tex]\bar X-1.96 \times {\frac{s}{\sqrt{n} } }[/tex] , [tex]\bar X +1.96 \times {\frac{s}{\sqrt{n} } }[/tex] )
= ( [tex]1110-1.96 \times {\frac{132}{\sqrt{1000} } }[/tex] , [tex]1110+1.96 \times {\frac{132}{\sqrt{1000} } }[/tex] )
= (1101.82 , 1118.18)
Therefore, 95% confidence interval for the population mean test score for high school seniors is (1101.82 , 1118.18).
