Question

An online survey of college parents was conducted. An email was sent to 41,000 parents who were listed in either of two college databases. Parents were invited to participate in the online survey. Out of those invited, 1,748 completed the online survey. The survey protected the anonymity of those participating in the survey but did not allow more than one response from an individual IP address.

One of the survey results was that 33% of mothers communicate at least once a day with their child while at school.

(a)

What was the response rate for this survey? (The response rate is the percentage of the planned sample—that is, those invited to participate—who responded. Round your answer to two decimal places.)

Use the quick method to estimate the margin of error for a random sample of size 1,748. (Round your answer to one decimal place.)

Do you think that the margin of error is a good measure of the accuracy of the survey's results? Explain your answer.

A) The margin of error is a good measure of the accuracy of the survey's results, because the study found that more than 25% of mothers communicate at least once a day with their child while at school.

B)The margin of error is a good measure of the accuracy of the survey's results, because the response rate was less than a quarter of the sample size.    

C)The margin of error is not a good measure of the accuracy of the survey's results, because the survey was sent via email and did not allow multiple responses from an individual IP address.

D)The margin of error is not a good measure of the accuracy of the survey's results, because we know nothing about the more than 95% of parents who did not respond.

Answer 1

The response rate for the survey = (total responses) / (total number of emails sent)

= 1748 / 41000

Response rate = 0.043 = 4.3%

This survey was to check for parents who communicate at least once a day with their child. Out of 1748 ,33% responded yes. We can use this as sample proportion = 33%.

This is an example of binomial distribution.

Margin of error = Standard error * Critical value

We can use normal approximation

Critical value at 5% = $Z_{5\%/2}$

= 1.96 ........using normal percentage tables with p = 0.025 = 2.5%

Standard error =

p(1-P) Vn

=

10.33(1 -0.33) T 1748

= 0.0112

Margin of error = 0.0112 * 1.96

MOE = 0.022 = 2.2%

Margin of error tells us how much % of error can be expected in the calculation. The error is expected because we calculate the parameters based on the sample and not the entire population. (Eg: out of 41000, 1748 responded. Therefore sample size=1748). Naturally, if there is an error we want it to be as small as possible. We would also like to minimize it).

A) The margin of error is a good measure of the accuracy of the survey's results, because the study found that more than 25% of mothers communicate at least once a day with their child while at school.

This is because out of the sample, 33% were found to communicate with children. Using MOE, we can say that 30.79% - 35.02% commuicate which is more than 25% even with the error. So it shows accuracy.