Summer 2017 Critical Thinking Assignment # 2 (50 points) INSTRUCTIONS: In this s
ID: 3257387 • Letter: S
Question
Summer 2017
Critical Thinking Assignment # 2 (50 points)
INSTRUCTIONS:
In this second assignment, you will further explore the data you obtained in critical assignment 1 by applying the concepts we have covered in chapters 6, 7, and 8.
USA
NON-USA
Mean GPA
3.23
3.63
Standard Deviation
2.64
3.57
Mean Age
24.50
29.50
Standard Deviation
7.88
10.33
Mean Hours spent on homework
9.44
10.73
Standard Deviation
6.76
7.68
Answer each of the questions below. Please show your calculations.
If one student is randomly selected from the USA born group, find the probability of getting someone with a GPA greater than 3.50.
If one student is randomly selected from the Non-USA born group, find the probability of getting someone with a GPA greater than 3.50.
If one student is randomly selected from the USA born group, find the probability of getting someone between the ages of 20 and 25.
If 9 students are randomly selected from the Non-USA born group, find the probability that their mean age is greater than 25 years.
If 25 students are randomly selected from the USA born group, find the probability that their mean GPA is between 2.50 and 3.50.
Calculate the age of a student at the 90th percentile for (a) the USA born group and (b) the foreign born group?
Assume that Broward College has announced that it will give a scholarship to students born in the USA who has a GPA between 3.75 and 4.00; calculate the percentage of students who will be eligible for a scholarship from the USA born group.
Assume that Broward College has announced that it will give scholarships to students who have GPA’s that falls within the top 2% of all GPA’s. Calculate the GPA a student will need to earn a scholarship from the USA born group.
In order to encourage students to spend more time doing homework at home, Broward College has announced that it will give a $100 scholarship to students who spend more than 10 hours per week doing homework at home. Calculate the percentage of students who will receive the scholarship from the NON-USA group.
Construct a 95% confidence interval estimate of the mean GPA for the USA born group. The sample size (n) for the USA born group is 231, Give your interpretation of the confidence interval estimate.
In a survey of 385 students at Broward College, North Campus, results revealed that 154 students were born outside of the USA and 231 were born in the USA. Construct a 90% confidence interval estimate for the proportion of students who were born outside the USA. Give your interpretation of the confidence interval estimate.
Suppose you wanted to estimate the percentage of students at North Campus who are foreign born, how many students must you survey if you wanted to be 95% confident in your result, with a margin of error of no more than 4%. Assume you have no estimate of p-hat.
Suppose you wanted to estimate the mean age of students born in the USA. How many students must you survey if you wanted to be 99% confident in your result and your error is no more than 3 years? Use the standard deviation from the table above.
In a survey of 385 students at Broward College, North Campus, results revealed that 154 students were born outside of the USA and 231 were born in the USA. Using these results, test the claim at the .03 significance level that the proportion of students who are born outside of the USA is less than 35 percent. Be sure to show all the steps in the hypothesis testing procedure including the correct wording of the final conclusion.
In a survey of 385 students at Broward College, North Campus, the mean GPA was found to be 3.15 with a standard deviation of .20. Using these results, test the claim at the .05 significance level that the mean GPA of students at North Campus is equal to 3.00. Be sure to show all the steps in the hypothesis testing procedure including the correct wording of the final conclusion.
USA
NON-USA
Mean GPA
3.23
3.63
Standard Deviation
2.64
3.57
Mean Age
24.50
29.50
Standard Deviation
7.88
10.33
Mean Hours spent on homework
9.44
10.73
Standard Deviation
6.76
7.68
Explanation / Answer
1.
We will consider that the GPA follows Normal distribution.
then if X=GPA
Xusa~N(3.23,2.642)
Then P(Xusa>3.5)=1-P(Xusa<3.5)= 1-0.54073=0.45927
2.Xnon-usa~N(3.63,3.572)
then, P(Xnon-usa>3.5)=1-P(Xnon-usa<3.5)= 1-0.48548=0.51452
3. We will consider that the age follows Normal distribution.
then if Y=Age
Yusa~N(24.5,7.882)
then, P(20<Yusa<25)=P(Yusa<25)-P(Ynon-usa<20)=0.52530-0.28398=
4.
The age mean of Non USA student = Ymean-non-usa
Ymean-non-usa ~N(29.50,(10.33/10)2 )=N(29.50,3.2662)
P(Ymean-non-usa>25)=1-P(Ymean-non-usa<25)=1-0.08413=0.91587
NOte: For calculation of Normal variable's probabilities w need to see a table or use any softwere like r or SAS.
According to the rule of this site, only first four parts have been solved.
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