I saw that this was already answered but I can't find it if anyone knows, can you point me in the right direction?
A recent poll of 700 people who work indoors found that 278 of them smoke.If the researchers want to be 98% confident of thier results to within 3.5%, how large a sample is necessary?
Here's one formula you can use for a problem of this type:
n = [(z-value)^2 * p * q]/E^2
Note: n = sample size needed; ^2 = squared; .4 for p and .6 for q (q = 1 - p). E = maximum error, which is .035 (3.5%) in the problem. Z-value is found using a z-table (find the value that corresponds to 98% confidence). Substitute into the formula. Round your answer to the next highest whole number.
Note: 278/700 is approximately .4 in decimal form.
I hope this will help.
2. A research team studied the ages of patients who had heart attacks caused by smoking. The ages of 34 patients who suffered a heart attack were as follows.
29 30 36 41 45 50 57 61 28 50 66 58
60 38 36 47 40 32 58 46 61 40 55 32
61 56 45 46 62 66 38 40 40 27
Construct a Stem-and-Leaf Plot using this data with Key: 5 | 7 = 57.
To determine the sample size necessary to estimate the proportion of smokers among people who work indoors with 98% confidence and within a margin of error of 3.5%, you can use the formula for sample size calculation:
n = (Z^2 * P * (1-P)) / E^2
Where:
n = sample size needed
Z = Z-score corresponding to the desired confidence level (in this case, 98% confidence corresponds to a Z-score of approximately 2.33)
P = estimated proportion (used 0.5 for a conservative estimate as we don't know the true proportion)
E = margin of error (0.035)
Plugging in the values:
n = (2.33^2 * 0.5 * (1-0.5)) / 0.035^2
Simplifying the equation:
n = (2.33^2 * 0.5 * 0.5) / 0.035^2
n = (5.4325 * 0.25) / 0.001225
n = 1.358125 / 0.001225
n = 1110.40
Rounding up to the nearest whole number, the sample size necessary to achieve the desired confidence level and margin of error is approximately 1111.