A survey in the U.K. interviewed shoppers encountered in grocery stores about wh

Question

A survey in the U.K. interviewed shoppers encountered in grocery stores about whether they had ever received injuries as a result of food or drink packaging, such as cuts sustained while cleaning up broken glass containers. Of the 200 who agreed to participate, 109 had received injuries "over the last few years" (27% of those injuries were singificant enough to be treated by a doctor or at an ER).

a. What is the best estimate, and 95% confidence interval for the proportion of shoppers in the population who have injured themselves?

b. Do you think this was a random sample of all UK consumers? What factors might have rendered it a non random sample?

Explanation / Answer

Proportion in the sample = 109 / 200

= 54.5%

Thus,

95% confidence interval =

0.545 +/- 1.96 *sqrt ( 0.545 * 0.456 / 200)

= ( 0.4759,0.6141)

Thus, proportion of injuries would be in the range = ( 47.59% to 61.41%)

The sample might not have been a random sample because of various biases that might have crept it. Including a response bias. Many people might not pay attention to minor injuries and often tend to forget about those.

Also, a sample taken from a single store might not be representative of the entire UK population.There might thus be an undercoverage bias as well.

Hope this helps.

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