Calculating 95% Confidence Interval for Mean Weight of Pretzel Bags
To find the 95% confidence interval for the mean weight of the pretzel bags, we can use the formula for the confidence interval for a mean when the population standard deviation is not known and the sample size is small (n < 30). Since we're dealing with a small sample size (n = 7), we'll use the t-distribution:
Confidence Interval (CI) = \(\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}\)
where:
- \(\bar{x}\) is the sample mean,
- \(t_{\alpha/2}\) is the t-value that corresponds to the \( \frac{\alpha}{2} \) percentile of the t-distribution with \(n - 1\) degrees of freedom,
- \(s\) is the sample standard deviation, and
- \(n\) is the sample size.
Given that:
- The sample mean, \(\bar{x}\), is 15.2 oz,
- The sample standard deviation, \(s\), is 0.7 oz, and
- The sample size, \(n\), is 7.
We need to find the t-value for a 95% confidence interval and 6 degrees of freedom (since \(n - 1 = 7 - 1 = 6\)). Typically, you would use a t-distribution table or statistical software to find this value. For a 95% confidence interval, we look up the value \(t_{0.025}\) since the two tails (alpha/2) of the curve each have 2.5% of the area.
Assuming a t-value of around 2.447 (you'll want to confirm this value using a t-distribution table or software), we can then calculate the confidence interval:
CI = \(15.2 \pm 2.447 \cdot \frac{0.7}{\sqrt{7}}\)
First calculate the margin of error (\(ME\)):
\(ME = 2.447 \cdot \frac{0.7}{\sqrt{7}}\)
\(ME = 2.447 \cdot \frac{0.7}{2.64575}\)
\(ME = 2.447 \cdot 0.2642\)
\(ME ≈ 0.6465\)
Now calculate the confidence interval bounds:
Lower bound = \(15.2 - 0.6465\)
Lower bound ≈ \(14.5535\)
Upper bound = \(15.2 + 0.6465\)
Upper bound ≈ \(15.8465\)
The 95% confidence interval is therefore approximately (14.5535, 15.8465).
Looking at the options provided:
a) 14.6 to 15.8
b) 13.4 to 17.0
c) 14.7 to 15.7
d) 14.4 to 16.0
The confidence interval we calculated is approximately 14.55 to 15.85, which makes option a) 14.6 to 15.8 the closest to our calculated interval and is thus the correct answer.
