# 8.4 Calculating the sample size n: continuous random variables and  (Page 2/13)

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 Required Sample Size (90%) Required Sample Size (95%) Tolerance Level 1691 2401 2% 752 1067 3% 271 384 5% 68 96 10%

This table is designed to show the maximum sample size required at different levels of confidence given an assumed p= 0.5 and q=0.5 as discussed above.

The acceptable error, called tolerance in the table, is measured in plus or minus values from the actual proportion. For example, an acceptable error of 5% means that if the sample proportion was found to be 26 percent, the conclusion would be that the actual population proportion is between 21 and 31 percent with a 90 percent level of confidence if a sample of 271 had been taken. Likewise, if the acceptable error was set at 2%, then the population proportion would be between 24 and 28 percent with a 90 percent level of confidence, but would require that the sample size be increased from 271 to 1,691. If we wished a higher level of confidence, we would require a larger sample size. Moving from a 90 percent level of confidence of a 95 percent level at a plus or minus 5% tolerance requires changing the sample size from 271 to 384. A very common sample size often seen reported in political surveys is 384. With the survey results it is frequently stated that the results are good to a plus or minus 5% level of “accuracy”.

Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.

From the problem, we know that the acceptable error, e , is 0.03 (3%=0.03) and ${z}_{\frac{\alpha }{2}}$ z 0.05 = 1.645 because the confidence level is 90%. The acceptable error, e , is the difference between the actual proportion p , and the sample proportion we expect to get from the sample.

However, in order to find n , we need to know the estimated (sample) proportion p ′. Remember that q ′ = 1 – p ′. But, we do not know p ′ yet. Since we multiply p ′ and q ′ together, we make them both equal to 0.5 because p q ′ = (0.5)(0.5) = 0.25 results in the largest possible product. (Try other products: (0.6)(0.4) = 0.24; (0.3)(0.7) = 0.21; (0.2)(0.8) = 0.16 and so on). The largest possible product gives us the largest n . This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size n , use the formula and make the substitutions.

$n=\frac{{z}^{2}{p}^{\prime }{q}^{\prime }}{{e}^{2}}$ gives $n=\frac{{1.645}^{2}\left(0.5\right)\left(0.5\right)}{{0.03}^{2}}=751.7$

Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.

## Try it

Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?

271 customers should be surveyed.Check the Real Estate section in your local

#### Questions & Answers

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