A university dean wishes to estimate the average number of hours his part-time instructors teach per week. The standard deviation is 2.6 hours. How large a sample must be selected if he wants to be 95% confident of finding whether the true mean differs from the sample mean by 1 hour?
To determine the sample size required, we can use the formula for estimating the sample size needed for a desired level of confidence interval.
The formula is given as:
n = [(Z * σ) / E]^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence (for a 95% confidence level, Z is 1.96)
σ = standard deviation
E = maximum error or difference between the sample mean and the true mean
In this case:
Z = 1.96 (for a 95% confidence level)
σ = 2.6 hours
E = 1 hour
Plugging in the values, we can calculate the required sample size (n):
n = [(1.96 * 2.6) / 1]^2
n = [5.096 / 1]^2
n = 25.96^2
n ≈ 676
Therefore, the sample size that must be selected to have a 95% confidence of finding whether the true mean differs from the sample mean by 1 hour is approximately 676.
To determine the sample size needed in this scenario, we can use the formula for sample size estimation for means with known standard deviation:
n = (Z * σ / E)^2
Where:
n = desired sample size
Z = Z-score for the desired level of confidence
σ = standard deviation
E = margin of error
In this case, the university dean wants to be 95% confident of finding whether the true mean differs from the sample mean by 1 hour. This means the confidence level (1 - α) is 0.95, which corresponds to a Z-score of 1.96 (obtained from a standard normal distribution table).
Plug in the given values into the formula:
n = (1.96 * 2.6 / 1)^2
Simplifying further:
n = (5.096 / 1)^2
n = (5.096)^2
n ≈ 25.964416
Since we cannot have a fraction of a person, it is prudent to round up the sample size to the nearest whole number. Therefore, the required sample size is approximately 26.
Hence, the university dean should select a sample size of 26 if he wants to be 95% confident of finding whether the true mean differs from the sample mean by 1 hour.
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Formula:
n = [(z-value * sd)/E]^2
...where n = sample size, z-value will be 1.96 using a z-table to represent the 95% confidence interval, sd = 2.6, E = 1, ^2 means squared, and * means to multiply.
Plug the values into the formula and finish the calculation. Round your answer to the next highest whole number.