Part II. (8 points) Imagine the engineering department at OSU is interested in u

Question

Part II. (8 points) Imagine the engineering department at OSU is interested in understanding the social habits of their students. Specifically, they would like to estimate the proportion of students who have a social networking profile. The provided bar chart was created from Student Information Survey for the Fall 2017 term, Fall ST314 Students with Social Networking Profile, n= 267 use this information to answer the following questions a. ( point) From the bar chart, what proportion of Fall ST314 not have a social networking profile? What b. point) Do you think it is appropriate to use the current ST314 Student Survey data to represent the population of all OSu C. (1 point) Check the sample size conditions for a confidence d. (2 points) Estimate the proportion of all OSU students with a e. (3 points) Interpret your estimate for p. Include context, and the point and interval estimates. engineering students? Why or why not? interval for p. Are these met? social networking profile. Use a confidence level of 90%. Show work No Yes

Explanation / Answer

Part II

(a) Proportion of students that don't have social networking profile = 11/267 = 0.0412

(b) No, it is not appropriate to use the current student survey data to represent the population of all OSU engineering students. The current ST314 students belong to only one class. The class has certain behavior so can't represent the entire enginnering student class

(c) Here sample size conditons np >= 5 , nq >= 5 and npq >= 15

so yes conditions are justified.

(d) proportion of students who use social networking profile p^ = 1 - 0.0412 = 0.9588

90% confidence interval = p^ +- Z90% sqrt [p^ (1-p^)/N]

= 0.9588 +- 1.645 * sqrt [0.0412 * 0.9588/267]

= 0.9588 + 1.645 * 0.0122

= (0.9387, 0.9789)

(e) here we can interpret our estimarte for p that there are only 95.88% students who use any social networking profile.

Here the interval estimate shows that there are 95% chance is there that the proportion of engineering stuents at OSU is in between 93.87 % to 97.89% .

Part III

(a) Here the actual proportion of all OSU studnets with a social profile is 0.69.But the confidence interval given above doesn't consist the value of 0.69 so it is not reasonable to assume the actual proportion of all OSU students with a social profile is not 0.69.

(b) Test statistic

Z = (p^ - 0.69)/ sqrt (0.69 * 0.31/267)

Z= (0.9588 - 0.69)/ 0.0283

Z = 9.50

so critical value of Z is Zcritical = 1.645

Z > Zcritical

so we shall reject the null hypothesis and can not conclude that actual proportion of all OSU studens with a social profile is 0.69.

(c) Yes, these students are techno-savy students which have more technical prowress then all Americans so they are more tend to use social accounts. so, 0.69 is a poor estimate for p.

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