Question

A survey was conducted to see if community college students preferred asynchronous distance learning (where learningoccurs through online channels without real-time interaction) or synchronous distance learning (where learning happens in real time, perhaps on Skype or Zoom).

Of the 250 students surveyed, 100 preferred asynchronous distance learning and 150 preferred synchronous distance learning. Construct a 95% confidence interval for the proportion of students who prefer asynchronous distance learning.

The local newspaper reported on the results of the survey by saying "over half of students prefer asynchronous distance learning". Is this statement supported by the confidence interval?

A) The 95% CI is (0.34,0.46). This interval does support the paper's statement.

B) The 95% CI is (0.34,0.46). This interval does not support the paper's statement.

C) The 95% CI is (0.54,0.66). This interval does not support the paper's statement.

D) The 95% CI is (0.54,0.66). This interval does support the paper's statement.

Next

Answer 1

The answer is:

B) The 95% CI is (0.34,0.46). This interval does not support the paper's statement.

One-Proportion Confidence Interval

We need to construct the 95% confidence interval for the population proportion. We have been provided with the following information: The sample size is N = 250, the number of favorable cases is X = 100 and the sample proportion is pˉ​=X/N​=100/250​=0.4, and the significance level is α=0.05 Critical Value Based on the information provided, the significance level is α=0.05, therefore the critical value is Zc​=1.96. This can be found by either using excel or the Z distribution table. Margin of Error The confidence interval: Therefore, based on the data provided, the 95% confidence interval for the population proportion is 0.3393<p<0.4607, which indicates that we are 95% confident that the true population proportion p is contained by the interval (0.3393,0.4607)