Mixing
Formulate the relevant hypotheses and test statistic (including its distribution)
$begingroup$
A manufacturer of space shuttle light bulbs claims that the defect rate of the bulbs is $0.1%$. You suspect the defect rate is actually higher, so you have checked $1000$ identical light bulbs from this manufacturer and found out that 3 of them are defect.
Formulate the relevant hypotheses and test statistic (including its distribution) and investigate the claim of the manufacturer at significance level $α = 0.05$.
I have problems to figure out which distribution it is, in my opinion, is a Poisson distribution but I'm not sure, can someone confirm it?
For the hypothesis, it's correct to assume that $H_0:mu=1$, $H_1:mu>1$?
Then how can I use the significance level with a Poisson distribution? (I know how to solve it but with a Normal) :(
Edit:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))$
For a $Bin(n,p)$ $P(X=k)=frac{n!}{k!(n-k)!}cdot p^kcdot(1-p)^{n-k}$
In order to calculate the probability, I have to substitute the values for $n$ and $p$ in the above equation and for each case the values of $k$ from $0$ to $2$.
Edit 2:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))= 1-0.981=0.019$ (from calculator), then how can I formulate a relevant hypothesis?
statistics
asked Jan 16 at 15:15
FTACFTAC
2649
$endgroup$
add a comment |
$begingroup$
A manufacturer of space shuttle light bulbs claims that the defect rate of the bulbs is $0.1%$. You suspect the defect rate is actually higher, so you have checked $1000$ identical light bulbs from this manufacturer and found out that 3 of them are defect.
Formulate the relevant hypotheses and test statistic (including its distribution) and investigate the claim of the manufacturer at significance level $α = 0.05$.
I have problems to figure out which distribution it is, in my opinion, is a Poisson distribution but I'm not sure, can someone confirm it?
For the hypothesis, it's correct to assume that $H_0:mu=1$, $H_1:mu>1$?
Then how can I use the significance level with a Poisson distribution? (I know how to solve it but with a Normal) :(
Edit:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))$
For a $Bin(n,p)$ $P(X=k)=frac{n!}{k!(n-k)!}cdot p^kcdot(1-p)^{n-k}$
In order to calculate the probability, I have to substitute the values for $n$ and $p$ in the above equation and for each case the values of $k$ from $0$ to $2$.
Edit 2:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))= 1-0.981=0.019$ (from calculator), then how can I formulate a relevant hypothesis?
statistics
asked Jan 16 at 15:15
FTACFTAC
2649
$endgroup$
3
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
1
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18
add a comment |
$begingroup$
A manufacturer of space shuttle light bulbs claims that the defect rate of the bulbs is $0.1%$. You suspect the defect rate is actually higher, so you have checked $1000$ identical light bulbs from this manufacturer and found out that 3 of them are defect.
Formulate the relevant hypotheses and test statistic (including its distribution) and investigate the claim of the manufacturer at significance level $α = 0.05$.
I have problems to figure out which distribution it is, in my opinion, is a Poisson distribution but I'm not sure, can someone confirm it?
For the hypothesis, it's correct to assume that $H_0:mu=1$, $H_1:mu>1$?
Then how can I use the significance level with a Poisson distribution? (I know how to solve it but with a Normal) :(
Edit:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))$
For a $Bin(n,p)$ $P(X=k)=frac{n!}{k!(n-k)!}cdot p^kcdot(1-p)^{n-k}$
In order to calculate the probability, I have to substitute the values for $n$ and $p$ in the above equation and for each case the values of $k$ from $0$ to $2$.
Edit 2:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))= 1-0.981=0.019$ (from calculator), then how can I formulate a relevant hypothesis?
statistics
asked Jan 16 at 15:15
FTACFTAC
2649
$endgroup$
A manufacturer of space shuttle light bulbs claims that the defect rate of the bulbs is $0.1%$. You suspect the defect rate is actually higher, so you have checked $1000$ identical light bulbs from this manufacturer and found out that 3 of them are defect.
Formulate the relevant hypotheses and test statistic (including its distribution) and investigate the claim of the manufacturer at significance level $α = 0.05$.
I have problems to figure out which distribution it is, in my opinion, is a Poisson distribution but I'm not sure, can someone confirm it?
For the hypothesis, it's correct to assume that $H_0:mu=1$, $H_1:mu>1$?
Then how can I use the significance level with a Poisson distribution? (I know how to solve it but with a Normal) :(
Edit:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))$
For a $Bin(n,p)$ $P(X=k)=frac{n!}{k!(n-k)!}cdot p^kcdot(1-p)^{n-k}$
In order to calculate the probability, I have to substitute the values for $n$ and $p$ in the above equation and for each case the values of $k$ from $0$ to $2$.
Edit 2:
$P(Xge3)=1-P(X<3)=1-(P(X=0)+P(X=1)+P(X=2))= 1-0.981=0.019$ (from calculator), then how can I formulate a relevant hypothesis?
statistics
statistics
asked Jan 16 at 15:15
FTACFTAC
2649
asked Jan 16 at 15:15
FTACFTAC
2649
edited Jan 16 at 22:01
FTAC
asked Jan 16 at 15:15
FTACFTAC
2649
asked Jan 16 at 15:15
FTACFTAC
2649
asked Jan 16 at 15:15
FTACFTAC
2649
2649
3
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
1
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18
add a comment |
3
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
1
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18
3
3
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
1
1
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
@lulu is correct that this is answered most directly by using the binomial distribution. Here is output (slightly edited for
relevance) from Minitab's exact
binomial test procedure.
Because this seems to be a homework problem, I will leave it to you
to interpret the P-value and to work out how it was
obtained from the distribution $mathsf{Binom}(n=1000,,p=0.001).$
Test for One Proportion
Test of p = 0.001 vs p > 0.001
Exact
Sample X N Sample p P-Value
1 3 1000 0.003000 0.080
Note: Your idea to use the Poisson approximation to the binomial distribution is not wrong. However, I suppose the exercise intends for you to use binomial.
The Poisson distribution that approximates $mathsf{Binom}(n=1000,,p=0.001)$ is $mathsf{Pois}(mu = 1).$ (The approximation is very good.) If $X$ has this Poisson distribution, you would expect to see one defective bulb: $E(X)=1.$ But you have seen three defective bulbs. That is more than expected; the question is whether it is enough more than expected to be called 'statistically significant' at the 5% level. Can you find $P(X ge 3)?$
The plot below compares the PDFs of $mathsf{Binom}(n=1000,,p=0.001)$ and $mathsf{Pois}(mu = 1).$
Addendum: From R: x = 0:4; pois.pdf = dpois(x,1); bino.pdf = dbinom(x, 1000, .001); returns (ignore line numbers in brackets):
cbind(x, pois.pdf, bino.pdf)
x pois.pdf bino.pdf
[1,] 0 0.36787944 0.36769542
[2,] 1 0.36787944 0.36806349
[3,] 2 0.18393972 0.18403174
[4,] 3 0.06131324 0.06128251
[5,] 4 0.01532831 0.01528996
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
$endgroup$
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
$begingroup$
From Minitab printout hypothesis and alternative are:Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.
$endgroup$
– BruceET
Jan 17 at 0:34
add a comment |
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$begingroup$
Hint:
@lulu is correct that this is answered most directly by using the binomial distribution. Here is output (slightly edited for
relevance) from Minitab's exact
binomial test procedure.
Because this seems to be a homework problem, I will leave it to you
to interpret the P-value and to work out how it was
obtained from the distribution $mathsf{Binom}(n=1000,,p=0.001).$
Test for One Proportion
Test of p = 0.001 vs p > 0.001
Exact
Sample X N Sample p P-Value
1 3 1000 0.003000 0.080
Note: Your idea to use the Poisson approximation to the binomial distribution is not wrong. However, I suppose the exercise intends for you to use binomial.
The Poisson distribution that approximates $mathsf{Binom}(n=1000,,p=0.001)$ is $mathsf{Pois}(mu = 1).$ (The approximation is very good.) If $X$ has this Poisson distribution, you would expect to see one defective bulb: $E(X)=1.$ But you have seen three defective bulbs. That is more than expected; the question is whether it is enough more than expected to be called 'statistically significant' at the 5% level. Can you find $P(X ge 3)?$
The plot below compares the PDFs of $mathsf{Binom}(n=1000,,p=0.001)$ and $mathsf{Pois}(mu = 1).$
Addendum: From R: x = 0:4; pois.pdf = dpois(x,1); bino.pdf = dbinom(x, 1000, .001); returns (ignore line numbers in brackets):
cbind(x, pois.pdf, bino.pdf)
x pois.pdf bino.pdf
[1,] 0 0.36787944 0.36769542
[2,] 1 0.36787944 0.36806349
[3,] 2 0.18393972 0.18403174
[4,] 3 0.06131324 0.06128251
[5,] 4 0.01532831 0.01528996
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
$endgroup$
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
$begingroup$
From Minitab printout hypothesis and alternative are:Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.
$endgroup$
– BruceET
Jan 17 at 0:34
add a comment |
$begingroup$
Hint:
@lulu is correct that this is answered most directly by using the binomial distribution. Here is output (slightly edited for
relevance) from Minitab's exact
binomial test procedure.
Because this seems to be a homework problem, I will leave it to you
to interpret the P-value and to work out how it was
obtained from the distribution $mathsf{Binom}(n=1000,,p=0.001).$
Test for One Proportion
Test of p = 0.001 vs p > 0.001
Exact
Sample X N Sample p P-Value
1 3 1000 0.003000 0.080
Note: Your idea to use the Poisson approximation to the binomial distribution is not wrong. However, I suppose the exercise intends for you to use binomial.
The Poisson distribution that approximates $mathsf{Binom}(n=1000,,p=0.001)$ is $mathsf{Pois}(mu = 1).$ (The approximation is very good.) If $X$ has this Poisson distribution, you would expect to see one defective bulb: $E(X)=1.$ But you have seen three defective bulbs. That is more than expected; the question is whether it is enough more than expected to be called 'statistically significant' at the 5% level. Can you find $P(X ge 3)?$
The plot below compares the PDFs of $mathsf{Binom}(n=1000,,p=0.001)$ and $mathsf{Pois}(mu = 1).$
Addendum: From R: x = 0:4; pois.pdf = dpois(x,1); bino.pdf = dbinom(x, 1000, .001); returns (ignore line numbers in brackets):
cbind(x, pois.pdf, bino.pdf)
x pois.pdf bino.pdf
[1,] 0 0.36787944 0.36769542
[2,] 1 0.36787944 0.36806349
[3,] 2 0.18393972 0.18403174
[4,] 3 0.06131324 0.06128251
[5,] 4 0.01532831 0.01528996
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
$endgroup$
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
$begingroup$
From Minitab printout hypothesis and alternative are:Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.
$endgroup$
– BruceET
Jan 17 at 0:34
add a comment |
$begingroup$
Hint:
@lulu is correct that this is answered most directly by using the binomial distribution. Here is output (slightly edited for
relevance) from Minitab's exact
binomial test procedure.
Because this seems to be a homework problem, I will leave it to you
to interpret the P-value and to work out how it was
obtained from the distribution $mathsf{Binom}(n=1000,,p=0.001).$
Test for One Proportion
Test of p = 0.001 vs p > 0.001
Exact
Sample X N Sample p P-Value
1 3 1000 0.003000 0.080
Note: Your idea to use the Poisson approximation to the binomial distribution is not wrong. However, I suppose the exercise intends for you to use binomial.
The Poisson distribution that approximates $mathsf{Binom}(n=1000,,p=0.001)$ is $mathsf{Pois}(mu = 1).$ (The approximation is very good.) If $X$ has this Poisson distribution, you would expect to see one defective bulb: $E(X)=1.$ But you have seen three defective bulbs. That is more than expected; the question is whether it is enough more than expected to be called 'statistically significant' at the 5% level. Can you find $P(X ge 3)?$
The plot below compares the PDFs of $mathsf{Binom}(n=1000,,p=0.001)$ and $mathsf{Pois}(mu = 1).$
Addendum: From R: x = 0:4; pois.pdf = dpois(x,1); bino.pdf = dbinom(x, 1000, .001); returns (ignore line numbers in brackets):
cbind(x, pois.pdf, bino.pdf)
x pois.pdf bino.pdf
[1,] 0 0.36787944 0.36769542
[2,] 1 0.36787944 0.36806349
[3,] 2 0.18393972 0.18403174
[4,] 3 0.06131324 0.06128251
[5,] 4 0.01532831 0.01528996
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
$endgroup$
Hint:
@lulu is correct that this is answered most directly by using the binomial distribution. Here is output (slightly edited for
relevance) from Minitab's exact
binomial test procedure.
Because this seems to be a homework problem, I will leave it to you
to interpret the P-value and to work out how it was
obtained from the distribution $mathsf{Binom}(n=1000,,p=0.001).$
Test for One Proportion
Test of p = 0.001 vs p > 0.001
Exact
Sample X N Sample p P-Value
1 3 1000 0.003000 0.080
Note: Your idea to use the Poisson approximation to the binomial distribution is not wrong. However, I suppose the exercise intends for you to use binomial.
The Poisson distribution that approximates $mathsf{Binom}(n=1000,,p=0.001)$ is $mathsf{Pois}(mu = 1).$ (The approximation is very good.) If $X$ has this Poisson distribution, you would expect to see one defective bulb: $E(X)=1.$ But you have seen three defective bulbs. That is more than expected; the question is whether it is enough more than expected to be called 'statistically significant' at the 5% level. Can you find $P(X ge 3)?$
The plot below compares the PDFs of $mathsf{Binom}(n=1000,,p=0.001)$ and $mathsf{Pois}(mu = 1).$
Addendum: From R: x = 0:4; pois.pdf = dpois(x,1); bino.pdf = dbinom(x, 1000, .001); returns (ignore line numbers in brackets):
cbind(x, pois.pdf, bino.pdf)
x pois.pdf bino.pdf
[1,] 0 0.36787944 0.36769542
[2,] 1 0.36787944 0.36806349
[3,] 2 0.18393972 0.18403174
[4,] 3 0.06131324 0.06128251
[5,] 4 0.01532831 0.01528996
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
edited Jan 16 at 21:22
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
answered Jan 16 at 17:41
BruceETBruceET
35.7k71440
35.7k71440
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
$begingroup$
From Minitab printout hypothesis and alternative are:Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.
$endgroup$
– BruceET
Jan 17 at 0:34
add a comment |
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
$begingroup$
From Minitab printout hypothesis and alternative are:Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.
$endgroup$
– BruceET
Jan 17 at 0:34
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
Thanks for the hint, I edited the question, please check it ;)
$endgroup$
– FTAC
Jan 16 at 20:33
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
$begingroup$
I put another edit, how can I formulate the hypothesis and investigate the significance level?
$endgroup$
– FTAC
Jan 16 at 22:01
1
1
$begingroup$
From Minitab printout hypothesis and alternative are:
Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.$endgroup$
– BruceET
Jan 17 at 0:34
$begingroup$
From Minitab printout hypothesis and alternative are:
Test of p = 0.001 vs p > 0.001. P-val is $P(X ge 3) approx 0.08,$ also from printout. Your calculator computation needs to be done again. Method OK; numbers wrong. Check them from Addendum.$endgroup$
– BruceET
Jan 17 at 0:34
add a comment |
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3
$begingroup$
The way the problem is phrased, binomial would appear to be more accurate (though the Poisson approximation should be fine). As the binomial is very easy to work with, I'd just do that.
$endgroup$
– lulu
Jan 16 at 15:23
1
$begingroup$
Edit seems on track. See addendum:
$endgroup$
– BruceET
Jan 16 at 21:18