A representative of a consumer advocacy group randomly selects 50 bags of a certain brand of chips and determines the net weight of each. He finds the mean of these selected bags to be 395 grams and the standard deviation to be 6.8 grams. Calculate a 90% confidence interval for the true mean of weight of chips. What is the lower limit of the interval?
Question
A representative of a consumer advocacy group randomly selects 50 bags of a certain brand of chips and determines the net weight of each. He finds the mean of these selected bags to be 395 grams and the standard deviation to be 6.8 grams. Calculate a 90% confidence interval for the true mean of weight of chips. What is the lower limit of the interval?
Solution
To calculate a 90% confidence interval for the true mean weight of the chips, we will use the formula for a confidence interval:
CI = x̄ ± Z * (σ/√n)
where:
- x̄ is the sample mean
- Z is the Z-score (which corresponds to the desired confidence level)
- σ is the standard deviation
- n is the sample size
For a 90% confidence interval, the Z-score is 1.645 (you can find this value in a standard Z-score table).
Let's plug in the given values:
CI = 395 ± 1.645 * (6.8/√50)
To find the lower limit of the interval, we will subtract the margin of error from the mean:
Lower limit = 395 - 1.645 * (6.8/√50)
Now, calculate the value inside the parentheses:
Lower limit = 395 - 1.645 * (6.8/7.071)
Lower limit = 395 - 1.645 * 0.961
Lower limit = 395 - 1.58
So, the lower limit of the 90% confidence interval for the true mean weight of the chips is approximately 393.42 grams.
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