Confidence interval for multinomial parameters
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I wonder how we can make confidence intervals for the parameters of a multinomial distribution. For example, suppose that we ask in a survey: what's your favourite hot drink: 1) Coffee 2) Thee 3) Chocolate? We got 100 answers: 23 for coffee 50 for Thee and 27 for chocolate. We would like to have confidences intervals for the proportion of the population in general.
We could separate the confidence intervals as if there were different binomials and use the normal approximation. For coffee, for a 95% confidence interval, we would find: $hat{p}pm z_alpha sqrt{frac{hat{p}({1-hat{p}})}{n}}$ with $hat{p}=frac{23}{100}$, $n=100$, $z_alpha=1.96$.
But I wonder if there is no better solution. In this example, we know that the sum of the proportions for coffee, thee and chocolate is exactly 1 and the previous calculation does not take this into account.
confidence-interval multinomial-coefficients
asked Dec 20 '18 at 11:01
olivierolivier
663
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add a comment |
$begingroup$
I wonder how we can make confidence intervals for the parameters of a multinomial distribution. For example, suppose that we ask in a survey: what's your favourite hot drink: 1) Coffee 2) Thee 3) Chocolate? We got 100 answers: 23 for coffee 50 for Thee and 27 for chocolate. We would like to have confidences intervals for the proportion of the population in general.
We could separate the confidence intervals as if there were different binomials and use the normal approximation. For coffee, for a 95% confidence interval, we would find: $hat{p}pm z_alpha sqrt{frac{hat{p}({1-hat{p}})}{n}}$ with $hat{p}=frac{23}{100}$, $n=100$, $z_alpha=1.96$.
But I wonder if there is no better solution. In this example, we know that the sum of the proportions for coffee, thee and chocolate is exactly 1 and the previous calculation does not take this into account.
confidence-interval multinomial-coefficients
asked Dec 20 '18 at 11:01
olivierolivier
663
$endgroup$
$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42
add a comment |
$begingroup$
I wonder how we can make confidence intervals for the parameters of a multinomial distribution. For example, suppose that we ask in a survey: what's your favourite hot drink: 1) Coffee 2) Thee 3) Chocolate? We got 100 answers: 23 for coffee 50 for Thee and 27 for chocolate. We would like to have confidences intervals for the proportion of the population in general.
We could separate the confidence intervals as if there were different binomials and use the normal approximation. For coffee, for a 95% confidence interval, we would find: $hat{p}pm z_alpha sqrt{frac{hat{p}({1-hat{p}})}{n}}$ with $hat{p}=frac{23}{100}$, $n=100$, $z_alpha=1.96$.
But I wonder if there is no better solution. In this example, we know that the sum of the proportions for coffee, thee and chocolate is exactly 1 and the previous calculation does not take this into account.
confidence-interval multinomial-coefficients
asked Dec 20 '18 at 11:01
olivierolivier
663
$endgroup$
I wonder how we can make confidence intervals for the parameters of a multinomial distribution. For example, suppose that we ask in a survey: what's your favourite hot drink: 1) Coffee 2) Thee 3) Chocolate? We got 100 answers: 23 for coffee 50 for Thee and 27 for chocolate. We would like to have confidences intervals for the proportion of the population in general.
We could separate the confidence intervals as if there were different binomials and use the normal approximation. For coffee, for a 95% confidence interval, we would find: $hat{p}pm z_alpha sqrt{frac{hat{p}({1-hat{p}})}{n}}$ with $hat{p}=frac{23}{100}$, $n=100$, $z_alpha=1.96$.
But I wonder if there is no better solution. In this example, we know that the sum of the proportions for coffee, thee and chocolate is exactly 1 and the previous calculation does not take this into account.
confidence-interval multinomial-coefficients
confidence-interval multinomial-coefficients
asked Dec 20 '18 at 11:01
olivierolivier
663
asked Dec 20 '18 at 11:01
olivierolivier
663
edited Dec 20 '18 at 11:16
olivier
asked Dec 20 '18 at 11:01
olivierolivier
663
asked Dec 20 '18 at 11:01
olivierolivier
663
asked Dec 20 '18 at 11:01
olivierolivier
663
663
$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42
add a comment |
$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42
$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42
$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42
add a comment |
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$begingroup$
Are you really interested in the individual or simultaneous confidence interval(s)? Your calculation above determines individual CIs whereby the preferred beverage proportion and the non-preferred proportion do sum to 1. The individual CIs would be >95% to obtain a simultaneous CI of 95%. Look up Tukey's method.
$endgroup$
– Phil H
Dec 20 '18 at 14:42