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Consider the following question:

A fast food manager wants to perform a statistical analysis of its restaurant customers’ weights. He collects a simple random sample of 81 customers. A 95% confidence interval based on this sample is (90 kg, 110 kg) which is based on a normal model for the mean.

A local newspaper claims that the weights of the customers at this fast food exceeds 85 kg. Is this claim supported by the confidence interval? Explain your reasoning.

I don't get the claim of the newspaper. Does it say that all customers must have weight above 85kg? I guess we can just argue about a population parameter, like mean, and not the range that values can scatter, right? because we don't know about the distribution of the values and it usually have a probability for any value, right? Please guide me over my understanding.

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  • $\begingroup$ Consider this as a testing problem where based on the confidence interval you have to accept or reject the hypothesis that the weights exceed 85 kg. $\endgroup$ – StubbornAtom Apr 19 at 18:20
  • $\begingroup$ @StubbornAtom so 85 is not in the interval and is less than it! what does it result! $\endgroup$ – Ahmad Apr 19 at 19:40
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If you make the additional assumption that weights (and not just the mean) are normally distributed, you can find the sample variance from the margin of error. Then use it, along with the estimated mean and your omnipresent z-table, to find Pr(W > 85). It's not perfect, but it's an estimate of the proportion of weighty customers.

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The newspaper could reasonably report that the AVERAGE weight exceeds 85 kg, although there is some uncertainty in that.

(I won't comment on the implausibility of the proposition that a fast-food manager weighed 81 customers.)

The sample mean is $100$ and the endpoints of the confidence interval would be $$ 100 \pm 1.96 \frac S{\sqrt{81}}, $$ so we have $$ 100 + 1.96\cdot \frac S 9 = 110 $$ and so $S = 9/1.96 \approx 4.592.$ If we take that to be the standard deviation, we have $$ \frac{85-100}{4.592} \approx -3.267 $$ The probability of being at least $3.267$ standard deviations below the mean is small. Somehow no children ever entered this fast-food establishment. Indeed, the plausibility of the numbers could bear some looking into. The more substantial issue is whether or how to take into account the uncertainty in estimation of the population standard deviation.

Your subject line is "Can I infer the range of a random variable based on a confidence interval for the mean?". Answering that, when taken literally, I'd have to say it cannot be done unless you also know the sample size. (But in this case we do.)

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  • $\begingroup$ could we say that with the probability of 95% the average customer weight is between 90 to 110, so the probability for 85 is very lower than 5%? $\endgroup$ – Ahmad Apr 19 at 19:43
  • $\begingroup$ @Ahmad : That is correct if confidence intervals are interpreted as probability intervals. That opens up some subtle questions, but in this case there's probably no problem with that. $\endgroup$ – Michael Hardy Apr 19 at 19:45
  • $\begingroup$ Could you please say if you used a hypothesis testing? What is your H0? What is your Ha, and what is your p_value? because I don't get your inference method here. $\endgroup$ – Ahmad Apr 20 at 20:53
  • $\begingroup$ @Ahmad : One can reject the hypothesis at the $5\%$ level if $85\text{ kg}$ is not within the $95\%$ confidence interval. $\endgroup$ – Michael Hardy Apr 21 at 2:37
  • $\begingroup$ 0.19% isn't exactly "microscopic". If a reporter makes one claim a day for 40 years, each with that probability of being wrong, they can expect 27 of those claims to be wrong. $\endgroup$ – Acccumulation Apr 22 at 21:28
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If we take "weights" to mean "average weight" and "supported" to mean "gives significant evidence for", then yes. The probability, given that the average weight is less that 85 kg, of seeing that confidence interval is very low. So seeing this evidence should cause significant updating towards the average weight being greater than 85 kg.

However, if "supports" means "fully justifies", then no. This does not conclusively prove the claim. 5% of the time, 95% confidence intervals do not contain the true mean. That's what "95%" means.

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