A class collects $218.43 for a field trip. If there are 27 students in the class, what is the average amount of money each stude
1 answer:
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Answer:
1) Null hypothesis:
Alternative hypothesis:
2) estimated proportion of people indicated that they watch the late evening news on this local CBS station
3)
And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"
4)
5)
And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"
6) We see that so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.
7) Null hypothesis:
Step-by-step explanation:
Data given and notation
n=100 represent the random sample taken
X represent the people indicated that they watch the late evening news on this local CBS station
estimated proportion of people indicated that they watch the late evening news on this local CBS station
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
Alternative hypothesis:
Part 2
estimated proportion of people indicated that they watch the late evening news on this local CBS station
Part 3
Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :
And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"
Part 4
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
Since we have all the info requires we can replace in formula (1) like this:
Part 5
Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :
And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"
Part 6
We see that so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.
Part 7
Null hypothesis:
Your answer simplified should be
