Question

suppose the investigators had made a rough guess of 320 for the value of s before collecting data. what sample size would be necessary to obtain an interval width of 45 ml for a confidence level of 95%? (round your answer up to the nearest whole number.)

Answer 1

To achieve a 95% confidence interval width of 45 ml with a standard deviation estimate of 320, the necessary sample size is 201 when rounding up to the nearest whole number.

Step-by-step explanation:

To calculate the sample size necessary to achieve a confidence interval width of 45 ml with a 95% confidence level and an estimated standard deviation (s) of 320, we use the formula for the margin of error (E) given by E = z * (s / sqrt(n)), where 'z' is the z-score corresponding to the confidence level and 'n' is the sample size.

The z-score for a 95% confidence level is approximately 1.96.

Setting the margin of error to half the desired interval width of 45 ml (i.e., 22.5 ml) allows us to solve for 'n' in n = (z * s / E)^2.

Plugging the values in gives us n = (1.96 * 320 / 22.5)^2, which calculates to a sample size of approximately 200.02.

We round up to the nearest whole number, so the necessary sample size would be 201.