The mass of garbage put out for weekly curbside pickup, per household, in a certain community, is reported to have a mean of 340 kg and a standard deviation of 85 kg. Based on this information:
a) Assume that the population is very large. For a sample of 200 households from tiles community, calculate, to the nearest %, the probability that the sample mean will be between 335 and 345 kg.
b) Redo Part (a), for a sample size of 1,000 households instead of 200.
c) Redo Part (a) (ie, n = 200), but this time assume a population of 400 households.
d) Assuming again that the population is very large, let us now suppose that 4 consecutive samples of 200 households are taken, and each one of them has a sample mean less than 330 kg. Calculate the probability of obtaining this result.
e) Based on your answer for Part (d), is there reason to believe that the mean weekly garbage output per household is not actually around 330 kg? If so, in which direction has the mean garbage output shifted? Explain your answers using your previous calculations for this question, plus your knowledge of statistics fundamentals.
I've attempted a) to d) assuming a normal distribution and using the z table, got the following results:
a) 59%
b) 94%
c) 72%
d)
Consider each 200 sample as a trial.
Let X be the number of trials with mean less than 330kg (# of success).
So X ~ Bin(4, 0.0485) (0.0485 obtained using normal distribution).
Have P(X=4) = 0%
But based on these results, I'm not sure how to answer e) I'm assuming they wanted us to say the mean has shifted but it seems like it's just around 330 based on my calculations.
Would appreciate any input or hints. Thanks!
Your answer to (a) must be incorrect, because if the population mean is presumed to be $\mu = 340$, for a sample mean $\bar X$ of any size, we must have $$\Pr[335 \le \bar X \le 340] < \Pr[-\infty < \bar X \le 340] = \frac{1}{2},$$ since $\bar X$ is normally distributed. At most half of the total probability mass of a normal distribution lies on one side of its mean. The correct calculation has, for $n = 200$, $$\bar X \sim \operatorname{Normal}\left(\mu = 340, \,\sigma = \frac{85}{\sqrt{200}}\right),$$ hence $$\Pr[335 \le \bar X \le 340] = \Pr\left[\frac{335 - 340}{85/\sqrt{200}} \le \frac{\bar X - \mu}{\sigma} \le \frac{340 - 340}{85/\sqrt{200}} \right] \approx \Pr[-0.83189 \le Z \le 0] \approx 0.297265.$$
I will not compute (b) but your answer is incorrect for the same reason that (a) is incorrect.
Again, (c) is incorrect for the same reason as (a). However, because the question assumes a finite population, it is important to note that a finite population correction is required for the standard error of the mean (i.e., the standard deviation of the sample mean): $$SE_{\text{fpc}} = \sigma \sqrt{\frac{N-n}{N-1}},$$ where $N = 400$ is the population size and $n = 200$ is the sample size, and $\sigma$ is the uncorrected standard deviation of $\bar X$.
For (d), assuming that $\mu = 340$ is the true population mean, consider the quantity $$\theta = \Pr[\bar X < 330].$$ Then the chance that four independent samples drawn from the population all have this outcome is $\theta^4$. It is not zero, but it is quite small.
For (e), the value you obtained in (d) is what we call a $p$-value. It is the probability that the outcome of an experiment is at least as extreme as what you observed, assuming the null hypothesis is true. In other words, if the true population mean were in fact $340$ kg, the chance that you would get four samples with means less than $330$ kg is extremely unlikely, so much so that your data would suggest that the true mean is not $340$ kg, but has decreased. In fact, this particular outcome is more rare than flipping a fair coin $17$ times in a row and getting all heads by random chance. If you were handed a coin, flipped it $17$ times, and got all heads, would you still think it was fair? It's not impossible, but it's extremely unlikely.