To investigate, ComputerWare asked 500 of its iMac purchasers what computer they

Question

To investigate, ComputerWare asked 500 of its iMac purchasers what computer they owned or had owned. Responses showed there were 83 new computer owners, 60 who had owned a Windows-based computer, and 357 who had owned a Macintosh. The results are graphed below.

The proportion (fraction) of first-time computer owners was 83/500 = 0.167. So, 0.167 is an estimate of the proportion of new computer owners. To determine the margin of error, compute the 95% confidence interval on the proportion. You should do this on your own using the technology of your choice to arrive at the same results shown below. Remember, 83/500 is a 'proportion', so it is easiest to use a z-interval proportion calculation.

ANSWER: The 95% confidence interval extends from 0.13 to 0.20. Therefore, it is likely that between 13% and 20% of iMac buyers are first-time computer owners.

Assumptions

The confidence interval on the proportion assumes that the sampling distribution of p is normal. The sampling distribution with N = 500 and the true portion (Pi) being 0.1667 can be found using the "Normal Approximation to the Binomial"

A second assumption is that each observation (customer) is independent of each other observation. There is no reason to suppose this assumption is violated.

Epilogue

After one year and the sale of 2,000,000 iMacs, Apple reports that one-third (0.333) of the new buyers of iMacs are first-time computer buyers. This is way outside the 95% confidence interval that we calculated which extends from 0.13 to 0.20.

Question: Do you think this is sampling error or might there be other factors at work here?

Explanation / Answer

You have provided the confidence interval range to be from 0.13 to 0.20

The value reported by Apple is 0.333

Although this does lie outside the confidence interval range, this does not mean that there is an error.

The sample of 500 was used to predict the population proportion value, and the interval is constructed with 95 % confidence. Means there is still 5% chance that the true proportion will lie outside this range.

The value of 0.333 is not a population proportion, it is still a sample proportion, with excessive large sample size.

Had this been a hypothesis testing problem, then the null hypothesis ( in this case, that the true proportion is 0.167) would have been rejected.

Hope this helps !

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