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Answer :
In language pertaining to the context of the problem, we can say:
Using methods that are correct 90% of the time, we estimate that the average weight of turkeys is between 14.0498 lbs and 16.6036 lbs.
To form a 90% confidence interval for the average weight of a turkey using the given data, we can use the following steps:
1. Calculate the sample mean:
Sum up all the turkey weights and divide by the total number of turkeys:
Mean = (19 + 21 + 15 + 14 + 12 + 20 + 10 + 18 + 12.5 + 15 + 13 + 12 + 15.4 + 18 + 16) / 15 ≈ 15.3267
2. Calculate the sample standard deviation:
Find the square root of the sum of squared deviations from the mean divided by (n-1):
Standard deviation = sqrt(((19-15.3267)^2 + (21-15.3267)^2 + ... + (16-15.3267)^2) / (15-1)) ≈ 2.9561
3. Calculate the margin of error:
The margin of error is determined by multiplying the critical value (z-score) by the standard deviation and dividing by the square root of the sample size. For a 90% confidence level, the critical value is approximately 1.645:
Margin of error = 1.645 * (2.9561 / sqrt(15)) ≈ 1.2769
4. Calculate the confidence interval:
The confidence interval is obtained by subtracting the margin of error from the sample mean and adding it to the sample mean:
Lower bound = Mean - Margin of error = 15.3267 - 1.2769 ≈ 14.0498
Upper bound = Mean + Margin of error = 15.3267 + 1.2769 ≈ 16.6036
with 90% confidence, we estimate that the mean weight of turkeys is between approximately 14.0498 lbs and 16.6036 lbs.
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