Question 1198701: Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time, , of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate . Assuming that the standard deviation of the population of shopping times at the supermarkets is 28 minutes, what is the minimum sample size she must collect in order for her to be 90% confident that her estimate is within 3 minutes of u ?
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Determine the Z-score for the desired confidence level:**
* For a 90% confidence level, the area in each tail of the standard normal distribution is (100% - 90%) / 2 = 5%.
* Find the z-score that corresponds to an area of 0.05 in the right tail of the standard normal distribution.
* Using a standard normal table or calculator, you'll find that z = 1.645.
**2. Calculate the Margin of Error (E)**
* E = Z * (σ / √n)
* where:
* E = desired margin of error (3 minutes)
* Z = z-score for the desired confidence level (1.645)
* σ = population standard deviation (28 minutes)
* n = sample size
**3. Rearrange the formula to solve for n:**
* n = (Z * σ / E)²
**4. Substitute the values and calculate the sample size:**
* n = (1.645 * 28 / 3)²
* n ≈ 182.34
**5. Round up to the nearest whole number:**
* Since we need a whole number of samples, round up n to 183.
**Therefore, the consultant must collect a minimum sample size of 183 shopping times to be 90% confident that her estimate of the mean shopping time is within 3 minutes of the true population mean.**
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