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Example 9.4 Estimating the Response to a New Sandwich Confidence Interval for a Proportion

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Objective To illustrate the procedure for finding a confidence interval for the proportion of customers who rate the new sandwich at least 6 on a 10-point scale.

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Background Information The fast food manager from Example 9.2 has already sampled 40 customers to estimate the population mean rating of its new sandwich. Recall that each rating is on a scale of 1 to 10, 10 being best. The manager would now like to use the same sample to estimate the proportion of customer who rate the sandwich at least 6. Her thinking is that these are the customers who are likely to purchase the sandwich again.

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Sandwich2.xls This file contains the solution. To create this output, we first count the number of ratings that are at least 6 in cell E6. The easiest way to do this is with the formula =COUNTIF(B4:B43,>=6) Then we calculate the sample proportion in cell E7 with the formula =E6/E3

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Results

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Results -- continued The rest is simply a matter of implementing the equations. Specifically, the standard error formula in cell E11 is =SQRT(E7*(1-E7)/E3) and the formula for the upper limit of the confidence interval, in cell E13 is =E7-E12*E11 Of course, the formula for the upper limit of the confidence interval is the same except with a plus sign.

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Results -- continued Then using the confidence interval limits p L = 0.475 and p U =0.775, we can check the assumption of the large sample size. With n=40, you can check that np L,n(1-p L ), np U, and n(1-p U ) are all well above 5, so that the validity of this confidence interval is established.

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9.19.1 | 9.2 | 9.3 | 9.5 | 9.6 | 9.7 | 9.8 | 9.9 | 9.10 | 9.11 | 9.12 | 9.13 | 9.14 | 9.159.29.39.59.69.79.89.99.109.119.129.139.149.15 Conclusions The output is fairly good news for the manager. Based on this sample of size 40, she can be 95% confident that the percentage of all customers who would rate the sandwich 6 or higher is somewhere between 47.5% and 77.5%. This is a large interval so there is a lot of uncertainty. To reduce this interval the manager would have to sample more customers.

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