One hundred five of the 195 members surveyed at a local gym say they worked out more often when they were in their twenties than they did when they were teenagers. Calculate the margin of error (ME), rounded to the nearest tenth of a percent. Is it reasonable that the federal health department claims the percentage for the entire country is 48%? Justify your answer.

Hint: ME equals plus or minus two times square root of start fraction p hat left parenthesis one minus p hat right parenthesis over N end fraction end square root, where p hat is the yes proportion in the sample and N is the sample size

No, it is not reasonable because the ME is 2.9% and the population proportion of 48% falls outside the sample interval of 53.8% ± 3.6%

Yes, it is reasonable because the ME is 7.1% and the population proportion of 54% falls within the sample interval of 48.0% ± 7.1%

Yes, it is reasonable because the ME is 7.1% and the population proportion of 48% falls within the sample interval of 53.8% ± 7.1%

No, it is not reasonable because the ME is 6.7% and the population proportion of 48% falls outside the sample interval of 53.8% ± 6.7%

Yes, it is reasonable, because the ME is 7.1% and 48% falls within the range of 53.8% to 7.1%.

Calculating the ME:

[tex]ME=\pm 2\sqrt{\frac{\hat{p}(1-\hat{p})}{N}}=\pm 2\sqrt{\frac{0.54(0.46)}{195}} =\pm 2\sqrt{\frac{0.2484}{195}}=0.071=7.1[/tex]

The sample proportion was 53.8%. This gives us
53.8% +/- 7.1%

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