PLEASE HELP 10 POINTS
1 answer:
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Answer:
a) The 95% confidence interval would be given by (509.592;550.308)
b) n=523 rounded up to the nearest integer
Step-by-step explanation:
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
s=70 represent the sample standard deviation
n=48 represent the sample size (variable of interest)
Confidence =95% or 0.995
Part a
The confidence interval for the mean is given by the following formula:
(1)
The degrees of freedom are df=n-1=48-1=47
Since the Confidence is 0.95 or 95%, 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(0.025,47)".And we see that
Now we have everything in order to replace into formula (1):
So on this case the 95% confidence interval would be given by (509.592;550.308)
Part b
The margin of error is given by this formula:
(a)
Assuming that
And on this case we have that ME =6msec, and we are interested in order to find the value of n, if we solve n from equation (a) we got:
(b)
The critical value for 95% of confidence interval is provided, , replacing into formula (b) we got:
So the answer for this case would be n=523 rounded up to the nearest integer