Question

In: Statistics and Probability

20. In a survey of 735 students at Broward College, North Campus, results revealed that 283...

20. In a survey of 735 students at Broward College, North Campus, results revealed that 283 students were foreign-born and 452 were native-born. Using these results, conduct a hypothesis test to test the claim, at the .03 significance level, that the proportion of students who are born outside of the USA is greater than 35 percent. Be sure to show all the steps in the hypothesis testing procedure including the null and alternative hypotheses, the calculation of the test statistic, and correct wording of the final conclusion.

21. In a survey of 735 students at Broward College, North Campus, the mean GPA was found to be 3.02 with a standard deviation of .42. 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 null and alternative hypotheses, the calculation of the test statistic, and correct wording of the final conclusion.

Solutions

Expert Solution

Solution:

Question 20

Here, we have to use z test for population proportion.

H0: p = 0.35 versus Ha: p > 0.35

This is an upper tailed or right tailed test.

We are given level of significance = α = 0.03

x = 283, n = 735

Sample proportion = P = x/n = 283/735 = 0.385034014

Test statistic = Z = (P – p)/sqrt(pq/n)

Where, P is sample proportion, p is population proportion, q = 1 – p, n is sample size.

q = 1 – 0.35 = 0.65

Z = (0.385034014 – 0.35) / sqrt(0.35*0.65/735)

Z = 1.9913

P-value = 0.0232

(by using z-table)

P-value < α = 0.03

So, we reject the null hypothesis

There is sufficient evidence to conclude that the proportion of students who are born outside of the USA is greater than 35 percent.

Question 21

One sample t test

H0: µ = 3.00 versus Ha: µ ≠ 3.00

Two tailed test

Given: Xbar = 3.02,

S = .42,

n = 735,

df = n – 1 = 734,

α = 0.05

Test statistic = t = (Xbar - µ)/[S/sqrt(n)]

t = (3.02 – 3.00)/[0.42/sqrt(735)]

t = 0.02/ 0.0155

t = 1.2910

P-value = 0.1971

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the mean GPA of students at North Campus is equal to 3.00.


Related Solutions

Is it true that the majority of North campus students text while driving? In a survey...
Is it true that the majority of North campus students text while driving? In a survey of 60 students at North Campus, 45 indicated that they often text while driving.   Using a .05 significance level, test the claim that the proportion of students at North Campus who text while driving is greater than 50%. Show all steps in the hypothesis testing procedure. Clearly enumerate the null and alternative hypotheses. Show calculation of test statistic and p-vale. Give wording of the...
Is it true that the majority of North campus students text while driving? In a survey...
Is it true that the majority of North campus students text while driving? In a survey of 60 students at North Campus, 45 indicated that they often text while driving.   Using a .05 significance level, test the claim that the proportion of students at North Campus who text while driving is greater than 50%. Show all steps in the hypothesis testing procedure. Clearly enumerate the null and alternative hypotheses. Show calculation of test statistic and p-vale. Give wording of the...
Is it true that the majority of North campus students text while driving? In a survey...
Is it true that the majority of North campus students text while driving? In a survey of 60 students at North Campus, 45 indicated that they often text while driving.   Using a .05 significance level, test the claim that the proportion of students at North Campus who text while driving is greater than 50%. Show all steps in the hypothesis testing procedure. Clearly enumerate the null and alternative hypotheses. Show calculation of test statistic and p-vale. Give wording of the...
Results of a survey revealed that the distribution of the amount of the monthly utility bill...
Results of a survey revealed that the distribution of the amount of the monthly utility bill of a 3-bedroom house using gas or electric energy had a mean of $97 and a standard deviation of $12. If the distribution of monthly bills can be considered mound-shaped and symmetric, what percentage of homes will have a monthly bill of more than $85 and less than $109? If nothing is known about the shape of the distribution of monthly bills, what percentage...
A check of dorm rooms on a large college campus revealed that 48% had refrigerators, 43%...
A check of dorm rooms on a large college campus revealed that 48% had refrigerators, 43% had TVs, and 63% had at least a TV or a refrigerator. What is the probability that a randomly selected dorm room has both a TV and a refrigerator? a: 0.00 b:0.28 c:0.41 d:0.59 e:0.72 What is the probability that a randomly selected dorm room has either a TV or a refrigerator, but not both? a: 0.00 b: 0.18 c: 0.35 d: 0.85 e:...
A student at a junior college conducted a survey of 20 randomly selected​ full-time students to...
A student at a junior college conducted a survey of 20 randomly selected​ full-time students to determine the relation between the number of hours of video game playing each​ week, x, and​ grade-point average, y. She found that a linear relation exists between the two variables. The​ least-squares regression line that describes this relation is ModifyingAbove y with caret equals negative 0.0559 x plus 2.9289. (a) Predict the​ grade-point average of a student who plays video games 8 hours per...
A student at a junior college conducted a survey of 20 randomly selected? full-time students to...
A student at a junior college conducted a survey of 20 randomly selected? full-time students to determine the relation between the number of hours of video game playing each? week, x, and? grade-point average, y. She found that a linear relation exists between the two variables. The? least-squares regression line that describes this relation is y = -0.0503x + 2.9381. (a) Predict the grade-point average of a student who plays video games 8 hours per week. The predicted grade-point average...
A student at a junior college conducted a survey of 20 randomly selected​ full-time students to...
A student at a junior college conducted a survey of 20 randomly selected​ full-time students to determine the relation between the number of hours of video game playing each​ week, x, and​ grade-point average, y. She found that a linear relation exists between the two variables. The​ least-squares regression line that describes this relation is ModifyingAbove y with caret equals negative 0.0592 x plus 2.9446. ​(a) Predict the​ grade-point average of a student who plays video games 8 hours per...
1: A survey questioned 1,000 high school students. The survey revealed that 46% are honor roll...
1: A survey questioned 1,000 high school students. The survey revealed that 46% are honor roll students. Of those who are honor roll students, 45% play sports in school and 21% of those that are not honor roll students, don't play sports. What is the probability that a high school student selected at random plays sports in school? 2: One of two small classrooms is chosen at random with equally likely probability, and then a student is chosen at random...
Cars on Campus. Statistics students at a community college wonder whether the cars belonging to students...
Cars on Campus. Statistics students at a community college wonder whether the cars belonging to students are, on average, older than the cars belonging to faculty. They select a random sample of 11 cars in the student parking lot and find the average age to be 7.5 years with a standard deviation of 5.6 years. A random sample of 20 cars in the faculty parking lot have an average age of 4.2 years with a standard deviation of 4 years....
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT