A random sample of 4000 U.S. citizens yielded 2150 who are in favor of outlawing cheesing. Find the point estimate for estimatin
1 answer:
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Answer:
d. Yes, because the confidence interval does not contain zero.
Step-by-step explanation:
We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.
The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.
Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;
P.Q. = ~
where, = sample mean SAT math score for in-state applicants = 540
= sample mean SAT math score for out-of-state applicants = 555
= sample standard deviation for in-state applicants = 20
= sample standard deviation for out-of-state applicants = 25
= sample of in-state applicants = 35
= sample of out-of-state applicants = 35
Also, =
= 22.64
<em>Here for constructing 95% confidence interval we have used Two-sample t test statistics.</em>
So, 95% confidence interval for the difference between population means () is ;
P(-1.997 < < 1.997) = 0.95 {As the critical value of t at 68 degree
of freedom are -1.997 & 1.997 with P = 2.5%}
P(-1.997 < < 1.997) = 0.95
P( <
<
) = 0.95
P( < (
) <
) = 0.95
<u>95% confidence interval for</u> () =
[ ,
]
=[,
]
= [-25.81 , -4.19]
Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].
This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.
So, option d is correct as Yes, because the confidence interval does not contain zero.