A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time student
1 answer:
Answer:
Step-by-step explanation:
The confidence interval for population mean is given by :-
Given : Sample size : , which is a large sample , so we apply z-test .
Sample mean :
Standard deviation :
Significance level :
Critical value :
Now, a confidence interval at the 95% level of confidence will be :-
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Answer:
b. 35.9, 39.1 ounces
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".
barX=37.465 represent the sample mean for the sample
population mean (variable of interest)
s=2.27 represent the sample standard deviation
n=10 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
barX +- t*[(s)/(sqrt{n))] (1)
In order to calculate the critical value t we need to find first the degrees of freedom, given by:
df=n-1=10-1=9
Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that t_(alpha/2)=2.26
Now we have everything in order to replace into formula (1):
37.465-2.26[(2.27)/(sqrt{10))]=35.9
37.465+2.26[(2.27)/(sqrt{10))]=39.1
So on this case the 95% confidence interval would be given by (35.9;39.1)
Average rate is given by,
for section A, a=1, b=3
hope this helps.