A sample of 100 cans of peas showed average weight of 14 ounces with standard deviation of 0.7 ounces. What is the value of margin of error? Using 95% confidence level
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A sample of 100 cans of peas showed average weight of 14 ounces with standard deviation of 0.7 ounces. What is the value of margin of error? Using 95% confidence level
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To calculate the margin of error (ME), we need to use the following formula:
ME = z * (σ / sqrt(n))
Where:
z is the z-score associated with the desired confidence level
σ is the population standard deviation (unknown, but we can estimate it using the sample standard deviation)
n is the sample size
In this case, we want a 95% confidence level, which means we need to use a z-score of 1.96 (this value can be obtained from a standard normal distribution table or calculator).
So, plugging in the values we have:
ME = 1.96 * (0.7 / sqrt(100))
ME = 1.96 * 0.07
ME = 0.1372 (rounded to four decimal places)
Therefore, the margin of error is approximately 0.1372 ounces. We can interpret this as follows: if we were to repeat the sampling process many times and construct a confidence interval for the population mean weight based on each sample, we would expect about 95% of those intervals to contain the true population mean weight within a range of ±0.1372 ounces.