To determine the standard deviation of the sampling distribution of the difference in means [tex]\(\bar{x}_R - \bar{x}_M\)[/tex], we can use the properties of standard deviations and sampling distributions.
Given:
- [tex]\(\bar{x}_R\)[/tex] is the mean popping time for a regular bag of microwave popcorn.
- [tex]\(\bar{x}_M\)[/tex] is the mean popping time for a mini bag of microwave popcorn.
- Mean popping time for regular bags, [tex]\(\mu_R = 140\)[/tex] seconds, with standard deviation [tex]\(\sigma_R = 20\)[/tex] seconds.
- Mean popping time for mini bags, [tex]\(\mu_M = 90\)[/tex] seconds, with standard deviation [tex]\(\sigma_M = 15\)[/tex] seconds.
- Sample size for regular bags, [tex]\(n_R = 25\)[/tex].
- Sample size for mini bags, [tex]\(n_M = 25\)[/tex].
The standard deviation of the sampling distribution of the difference in sample means ([tex]\(\bar{x}_R - \bar{x}_M\)[/tex]) is given by the formula:
[tex]\[
\sigma_{\bar{x}_R - \bar{x}_M} = \sqrt{\frac{\sigma_R^2}{n_R} + \frac{\sigma_M^2}{n_M}}
\][/tex]
Let's plug in the given values:
1. Calculate the variance for the regular bags of popcorn:
[tex]\[
\frac{\sigma_R^2}{n_R} = \frac{20^2}{25} = \frac{400}{25} = 16
\][/tex]
2. Calculate the variance for the mini bags of popcorn:
[tex]\[
\frac{\sigma_M^2}{n_M} = \frac{15^2}{25} = \frac{225}{25} = 9
\][/tex]
3. Sum these variances to get the variance of the difference in sample means:
[tex]\[
16 + 9 = 25
\][/tex]
4. Take the square root of this sum to find the standard deviation of the difference in sample means:
[tex]\[
\sigma_{\bar{x}_R - \bar{x}_M} = \sqrt{25} = 5
\][/tex]
Therefore, the standard deviation of the sampling distribution of [tex]\(\bar{x}_R - \bar{x}_M\)[/tex] is [tex]\(5\)[/tex] seconds. The correct answer is:
5 seconds
