Car m left town q and car n left town p at the same time. at noon car m traveled 2/3 of the distannce from q to p and car n had
1 answer:
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Answer:
a)
So on this case the 99% confidence interval would be given by (13.83;28.17)
b)
c)
d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.
Step-by-step explanation:
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".
Part a
represent the sample mean
population mean (variable of interest)
represent the population standard deviation
n=350 represent the sample size
99% confidence interval
The confidence interval for the mean is given by the following formula:
(1)
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: "=-NORM.INV(0.005,0,1)".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 (13.83;28.17)
Part b
The margin of error is given by:
Part c
The margin of error is given by:
Part d
As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:
D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.
9(2w−y)=21w−9y
First you multiply the numbers in the parenthesis by 9
subtract. 21w-18w=3w
add. -9y+9y=0
divide.
Final answer is 0.