Recently the U.S. Department of Education released a report on online learning s

Question

Recently the U.S. Department of Education released a report on online learning stating that blended instruction, a combination of conventional face-to-face and online instruction, appears more effective in terms of student performance than conventional teaching. You decide to poll the incoming students at your institution to see if they prefer courses that blend face-to-face instruction with online components. In an SRS of 300 incoming students, you find that 214 prefer this type of course (a) What is the sample proportion who prefer this type of blended instruction? (Round your answer to two decimal places.) (b) If the population proportion for all students nationwide is 89%, what is the standard deviation of p? (Round your answer to four decimal places.) (c) Using the 68-95-99.7 rule, if you had drawn ar SRS from the United States, you would expect p to fall between what two proportions about 95% of the time? (Round your answers to four decimal places.) s) (d) Based on your result in part (a), do you think that the incoming students at your institution prefer this type of instruction more, less, or about the same as students nationally? Explain your answer The sample proportion was not in the interval thus we should believe that our students prefer biended instruction more than students nationaly The sample proportion was equal to the national average, 1hus we should believe that our students prefer blended instruction about the same as students nationally The sample proportion was n the mterval, thus we should believe that our students prefer blended instruction less than students nationally The sample proportion was not in the interval thus we should believe that our students prefer blended instruction less than students nationaly The sample proportion was in the interval, thus we should believe that our students prefer blended instruction more than students nationally

Explanation / Answer

a. Sample Proportion for those who prefer = 214/300 = 0.7133

b. Standard deviation of p^ = root [p(1-p)/n] where p is the proportion who prefers blended instructions and n is the number of students in the sample.

So, SD of p^ = root [ (0.89*0.11)/300 ] = 0.018

c. For 65% - 95% - 99.7% range, we need z values at these confidence intervals. As this is a two tail test, the z values would be at alpha by 2 values of level of signifance.

Now, we know that 68% confidence interval means 1 standard deviation around the mean, 95% means 2 standard deviations and 99.7% of confidence interval means 3 standard deviations around the mean.

So, the range required for 95% CI will be = 0.89 - 2*0.018 and 0.89 + 2*0.018 = 0.854 < x < 0.926

d. Option 4 because our proportion of 0.7133 in the sample who prefer the blended instruction is less than the population range at 95% CI or 2 standard deviations, hence the sample's students prefer blended instructions less.

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