Question 1088075: Please help me solve statistics question.
A sample of students from an Introductory Marketing class was polled regarding the number of hours they spent studying for the last statistics exam. All students anonymously submitted the number of hours on a 3 by 5 card. There were 24 individuals in the one section of the course polled. The data was used to make inferences regarding the other students taking the course. There data are below:
5
2
17
15
9
10
3
18
13
7
12
23
15
19
15
12
8
6
8
7
8
13
21
18
Compute a 95% confidence interval of the true population proportion of marketing students who spent less than 10 hours studying for the last statistics exam. Interpret your answer.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! There are 10 students in the 24 student sample that spent less than 10 hours studying for the final
:
sample proportion is 10/24 = 5/12 approx 0.42
:
We want a 95% confidence interval of the true population proportion
:
Standard error(SE) = square root( (0.41) * (1-0.41) / 24 ) = 0.1004
:
Margin of Error(ME) = critical value(CV) * SE
:
The CV will be expressed as a t statistic(t*), since the sample size(24) < 30
:
Alpha(a) = 1 - (95/100) = 0.05
:
The critical probability(p*) = 1 - (a/2) = 0.975
:
The degrees of freedom(DF) = 24 -1 = 23
:
t* is the t statistic having DF = 23 and p* = 0.975
:
t* = 2.069
:
ME = 2.069 * 0.1004 = 0.2077 approx 0.21
:
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The 95% confidence interval is 0.42 + or - 0.21 = (0.21, 0.63)
:
if we used the same sampling method to select different samples and computed a 95% confidence interval estimate for each sample, we would expect the true population parameter to fall within the interval estimates 95% of the time.
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