To determine the shape of the sampling distribution of the difference in mean popping times, [tex]\(\bar{x}_R - \bar{x}_M\)[/tex], for regular-sized and mini-sized microwave popcorn bags, we need to consider several key factors.
1. Distribution of Individual Popping Times:
- The popping times for regular bags ([tex]\(R\)[/tex]) are Normally distributed with a mean ([tex]\(\mu_R\)[/tex]) of 140 seconds and a standard deviation ([tex]\(\sigma_R\)[/tex]) of 20 seconds.
- The popping times for mini bags ([tex]\(M\)[/tex]) are Normally distributed with a mean ([tex]\(\mu_M\)[/tex]) of 90 seconds and a standard deviation ([tex]\(\sigma_M\)[/tex]) of 15 seconds.
2. Sampling and Central Limit Theorem:
- Independent random samples are taken, each of size 25 ([tex]\(n_R = n_M = 25\)[/tex]).
- According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately Normally distributed if the sample size is sufficiently large. Typically, a sample size of 25 is considered sufficiently large.
3. Distribution of Sample Means:
- Let [tex]\(\bar{x}_R\)[/tex] be the mean popping time for the sample of regular bags, and [tex]\(\bar{x}_M\)[/tex] be the mean popping time for the sample of mini bags.
- The distribution of [tex]\(\bar{x}_R\)[/tex] and [tex]\(\bar{x}_M\)[/tex] will each be approximately Normally distributed due to the Central Limit Theorem, as:
- [tex]\(\bar{x}_R \sim N(\mu_R, \frac{\sigma_R^2}{n_R})\)[/tex]
- [tex]\(\bar{x}_M \sim N(\mu_M, \frac{\sigma_M^2}{n_M})\)[/tex]
4. Distribution of the Difference in Sample Means:
- The difference in sample means, [tex]\(\bar{x}_R - \bar{x}_M\)[/tex], will also be Normally distributed. This results from the fact that the difference of two Normally distributed random variables is also Normally distributed.
- The mean and variance of the difference in sample means are given by:
[tex]\[ E(\bar{x}_R - \bar{x}_M) = \mu_R - \mu_M \][/tex]
[tex]\[ \text{Var}(\bar{x}_R - \bar{x}_M) = \frac{\sigma_R^2}{n_R} + \frac{\sigma_M^2}{n_M} \][/tex]
- Therefore, [tex]\(\bar{x}_R - \bar{x}_M \sim N(\mu_R - \mu_M, \sqrt{\frac{\sigma_R^2}{n_R} + \frac{\sigma_M^2}{n_M}})\)[/tex]
Based on the above points:
- Since both the individual popping time distributions for the regular and mini bags are Normal and we are considering the mean of sufficiently large samples (sample size 25 for each), the shape of the sampling distribution of [tex]\(\bar{x}_R - \bar{x}_M\)[/tex] is best described as Normal due to the Central Limit Theorem and the properties of Normal distributions.
Thus, the correct answer is:
Normal because both population distributions are Normal
