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, use this information to answer the following questions a. (1 point) From the bar chart, what proportion of Fall ST314 Fall ST314 Students with Social Networking Profile, n 267 256 students do not have a social networking profile? What 8 proportion do? (1 point) Do you think it is appropriate to use the current ST314 Student Survey data to represent the population of all OSU . engineering students? Why or why not? b. 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 interval for p. Are these met? No Yes social networking profile. Use a confidence level of 90%. Show work

Explanation / Answer

a) Sample size = n = 267

number of students who do not have a social networking profile = 11

therefore proportion of Fall ST214 students  do not have a social networking profile = 11 / 267 = 0.0412

This is less than 0.05 so it is unusual proportion.

b) Yes because the sample size is very large.

c) If n * p > 10 and n * ( 1 - p) > then we can used this sample size for confidence interval.

here n*p = 11> 10 and n* ( 1 - p) = 256 > 10

So we can used this sample size for finding confidence interval for population proportion.

d) Here we want to estimate 90% confidence interval for population proportion

We can used one sample proportion z interval.

Let' write given information.

n = sample size = 267

x = number of success = 11.

level of confidence = 90%

Using minitab.

The command for one sample proportion z test in minitab is

Stat>>>Basic statistics>>>1-proportion...

Then click on summarized data

number of events = x = 11

Number of trials = n = 267

then click on option

Level of confidence in percentage = c = 90%

so put "Confidence level " = 90

Alternative = not equal

then click on "Use test and interval based on normal approximation"

Then click on OK and again click on OK

So we get the following output

Test and CI for One Proportion

Sample X N Sample p 90% CI
1 11 267 0.041199 (0.021192, 0.061205)

From the above output the 90% confidence interval for population proportion is (0.021192, 0.061205)

e) This confidence interval capture about 90% of the times the true value of population proportion of the students who do not have a social networking profile.

That is 90% of the times the true proportion lies between 0.021192 to 0.061205

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