You measure 49 dogs' weights, and find they have a mean weight of 30 ounces. Assume the population standard deviation is 12.2 ounces.
Based on this, what is the margin of error associated with a 90% confidence interval for the true population mean dog weight.
To find the margin of error for a 90% confidence interval, we can use the formula:
Margin of Error = Z * (Standard Deviation / sqrt(n))
where:
Z represents the z-score associated with the desired level of confidence (in this case, 90%)
Standard Deviation is the population standard deviation (12.2 ounces)
n is the sample size (49)
First, let's find the Z-score for a 90% confidence level. The Z-score can be found using a standard normal distribution table or a calculator. For a z-score corresponding to a 90% confidence level, the value is approximately 1.645.
Now we can substitute the given values into the formula:
Margin of Error = 1.645 * (12.2 / sqrt(49))
Calculating the square root of 49 gives us:
Margin of Error = 1.645 * (12.2 / 7)
Finally, solving the equation:
Margin of Error ≈ 2.84 ounces
Therefore, the margin of error associated with a 90% confidence interval for the true population mean dog weight is approximately 2.84 ounces.