Jeremy boiled one cup of water and then let it cool at room temperature. The table shows the temperature, in degrees Fahrenheit,
1 answer:
You might be interested in
Answer:
a) The 99% confidence interval would be given by (5004.168;12603.832)
b) For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
represent the sample mean for the sample
population mean (variable of interest)
s=5629 represent the sample standard deviation
n=20 represent the sample size
2) Part a
The confidence interval for the mean is given by the following formula:
(1)
In order to calculate the critical value we need to find first the degrees of freedom, given by:
Since the Confidence is 0.99 or 99%, the value of and
, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,19)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 99% confidence interval would be given by (5004.168;12603.832)
3) Part b
For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I
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.