Can I infer the range of a random variable based on a confidence interval for the mean?
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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.
confidence-interval
asked 3 hours ago
AhmadAhmad
1137
$endgroup$
add a comment |
$begingroup$
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.
confidence-interval
asked 3 hours ago
AhmadAhmad
1137
$endgroup$
$begingroup$
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
2 hours ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago
add a comment |
$begingroup$
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.
confidence-interval
asked 3 hours ago
AhmadAhmad
1137
$endgroup$
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.
confidence-interval
confidence-interval
asked 3 hours ago
AhmadAhmad
1137
asked 3 hours ago
AhmadAhmad
1137
edited 2 hours ago
Ahmad
asked 3 hours ago
AhmadAhmad
1137
asked 3 hours ago
AhmadAhmad
1137
asked 3 hours ago
AhmadAhmad
1137
1137
$begingroup$
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
2 hours ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago
add a comment |
$begingroup$
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
2 hours ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Mike AndersonMike Anderson
1513
$endgroup$
add a comment |
$begingroup$
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
$$
100 + 1.96cdot frac S 9 = 110
$$
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 microscopic. 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.)
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered 2 hours ago
Mike AndersonMike Anderson
1513
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Mike AndersonMike Anderson
1513
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Mike AndersonMike Anderson
1513
$endgroup$
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.
answered 2 hours ago
Mike AndersonMike Anderson
1513
answered 2 hours ago
Mike AndersonMike Anderson
1513
answered 2 hours ago
Mike AndersonMike Anderson
1513
answered 2 hours ago
Mike AndersonMike Anderson
1513
1513
add a comment |
add a comment |
$begingroup$
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
$$
100 + 1.96cdot frac S 9 = 110
$$
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 microscopic. 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.)
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
add a comment |
$begingroup$
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
$$
100 + 1.96cdot frac S 9 = 110
$$
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 microscopic. 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.)
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
add a comment |
$begingroup$
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
$$
100 + 1.96cdot frac S 9 = 110
$$
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 microscopic. 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.)
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
$endgroup$
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
$$
100 + 1.96cdot frac S 9 = 110
$$
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 microscopic. 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.)
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
answered 1 hour ago
Michael HardyMichael Hardy
4,0551430
4,0551430
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
add a comment |
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
@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
1 hour ago
add a comment |
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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
2 hours ago
$begingroup$
@StubbornAtom so 85 is not in the interval and is less than it! what does it result!
$endgroup$
– Ahmad
1 hour ago