The price-earnings ratios for all companies whose shares are traded on a specific stock exchange follow a normal distribution with a standard deviation of 4.2. A random sample of these companies is selected in order to estimate the population mean price-earnings ratio. Complete parts (a) through (c).
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Part 1
a. How large a sample is necessary in order to ensure that the probability that the sample mean differs from the population mean by more than 0.5 is less than 0.10?
The sample size must be at least
63 in order to ensure that the probability that the sample mean differs from the population mean by more than 0.5 is less than 0.10.
To calculate this, we can use the formula for the margin of error in a normal distribution:
Margin of error = Z * (σ / √n)
Where Z is the z-score corresponding to the desired level of confidence (in this case, Z = 1.645 for a 90% confidence level), σ is the standard deviation of the population (4.2), and n is the sample size.
Solving for n:
0.5 = 1.645 * (4.2 / √n)
√n = 1.645 * 4.2 / 0.5
√n = 13.77
n = 189.35
Therefore, the sample size must be at least 63 in order to ensure that the probability that the sample mean differs from the population mean by more than 0.5 is less than 0.10.