Mixing
Investigate the significance level $α = 0.01$.
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value.
For statistical purposes, you decide to throw $4$ fifty-cent coins in this machine and let $overline{X}_4$ be the estimator for the mean weight $μ$ of a fifty-cent coin.
Assume that a single measurement has a normal distribution with expectation $μ$ and variance $σ^2 = 0.04$.
You find out that $overline{x}_4 = 7.54$.
Investigate by using a statistical test whether $μ$ equals $7.5$ at significance level $α = 0.01$.
Solve:
I assume $H_0:mu=7.5$ and $H_1:muneq7.5$
So it is a two-tailed problem and I have an $alpha=0.005$ on the right and the same on the left. So from the table the critical values for these $alpha$ are 2.57 and -2.57.
So I can compute $Z=frac{7.48-7.5}{frac{sqrt{0.004}}{sqrt{4}}}=-0.63$ that is inside the region where we don't reject $H_0$ so we don't reject it.
Is it correct? Can someone please help me? Or I should use the p-value?
probability statistics
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value.
For statistical purposes, you decide to throw $4$ fifty-cent coins in this machine and let $overline{X}_4$ be the estimator for the mean weight $μ$ of a fifty-cent coin.
Assume that a single measurement has a normal distribution with expectation $μ$ and variance $σ^2 = 0.04$.
You find out that $overline{x}_4 = 7.54$.
Investigate by using a statistical test whether $μ$ equals $7.5$ at significance level $α = 0.01$.
Solve:
I assume $H_0:mu=7.5$ and $H_1:muneq7.5$
So it is a two-tailed problem and I have an $alpha=0.005$ on the right and the same on the left. So from the table the critical values for these $alpha$ are 2.57 and -2.57.
So I can compute $Z=frac{7.48-7.5}{frac{sqrt{0.004}}{sqrt{4}}}=-0.63$ that is inside the region where we don't reject $H_0$ so we don't reject it.
Is it correct? Can someone please help me? Or I should use the p-value?
probability statistics
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
$endgroup$
add a comment |
$begingroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value.
For statistical purposes, you decide to throw $4$ fifty-cent coins in this machine and let $overline{X}_4$ be the estimator for the mean weight $μ$ of a fifty-cent coin.
Assume that a single measurement has a normal distribution with expectation $μ$ and variance $σ^2 = 0.04$.
You find out that $overline{x}_4 = 7.54$.
Investigate by using a statistical test whether $μ$ equals $7.5$ at significance level $α = 0.01$.
Solve:
I assume $H_0:mu=7.5$ and $H_1:muneq7.5$
So it is a two-tailed problem and I have an $alpha=0.005$ on the right and the same on the left. So from the table the critical values for these $alpha$ are 2.57 and -2.57.
So I can compute $Z=frac{7.48-7.5}{frac{sqrt{0.004}}{sqrt{4}}}=-0.63$ that is inside the region where we don't reject $H_0$ so we don't reject it.
Is it correct? Can someone please help me? Or I should use the p-value?
probability statistics
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
$endgroup$
If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value.
For statistical purposes, you decide to throw $4$ fifty-cent coins in this machine and let $overline{X}_4$ be the estimator for the mean weight $μ$ of a fifty-cent coin.
Assume that a single measurement has a normal distribution with expectation $μ$ and variance $σ^2 = 0.04$.
You find out that $overline{x}_4 = 7.54$.
Investigate by using a statistical test whether $μ$ equals $7.5$ at significance level $α = 0.01$.
Solve:
I assume $H_0:mu=7.5$ and $H_1:muneq7.5$
So it is a two-tailed problem and I have an $alpha=0.005$ on the right and the same on the left. So from the table the critical values for these $alpha$ are 2.57 and -2.57.
So I can compute $Z=frac{7.48-7.5}{frac{sqrt{0.004}}{sqrt{4}}}=-0.63$ that is inside the region where we don't reject $H_0$ so we don't reject it.
Is it correct? Can someone please help me? Or I should use the p-value?
probability statistics
probability statistics
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
asked Jan 27 at 8:33
Mark JaconMark Jacon
1127
1127
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have written the population variance as $sigma^2 = 0.004,$ not $0.04.$
Also, in the problem you state that $bar X = 7.54,$ but in your computation you have $7.48.$
Otherwise, you seem to be on the right track.
After fixing your typos, you should have $Z = 0.4.$ Then at level $alpha = 0.01 = 1%,$
you would reject if $|Z| > 2.576,$ so you cannot reject.
The P-value
is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.
My computations from R statistical software are shown below:
(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565
Also, Minitab statistical software gives the following relevant output:
One-Sample Z
Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2
N Mean SE Mean 99% CI Z P
4 7.540 0.100 (7.282, 7.798) 0.40 0.689
Because your hypothetical mean $mu = 7.5$ is included in the 99%
confidence interval, you would not reject at the 1% level.
answered Jan 27 at 10:42
BruceETBruceET
36k71540
$endgroup$
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You have written the population variance as $sigma^2 = 0.004,$ not $0.04.$
Also, in the problem you state that $bar X = 7.54,$ but in your computation you have $7.48.$
Otherwise, you seem to be on the right track.
After fixing your typos, you should have $Z = 0.4.$ Then at level $alpha = 0.01 = 1%,$
you would reject if $|Z| > 2.576,$ so you cannot reject.
The P-value
is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.
My computations from R statistical software are shown below:
(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565
Also, Minitab statistical software gives the following relevant output:
One-Sample Z
Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2
N Mean SE Mean 99% CI Z P
4 7.540 0.100 (7.282, 7.798) 0.40 0.689
Because your hypothetical mean $mu = 7.5$ is included in the 99%
confidence interval, you would not reject at the 1% level.
answered Jan 27 at 10:42
BruceETBruceET
36k71540
$endgroup$
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
add a comment |
$begingroup$
You have written the population variance as $sigma^2 = 0.004,$ not $0.04.$
Also, in the problem you state that $bar X = 7.54,$ but in your computation you have $7.48.$
Otherwise, you seem to be on the right track.
After fixing your typos, you should have $Z = 0.4.$ Then at level $alpha = 0.01 = 1%,$
you would reject if $|Z| > 2.576,$ so you cannot reject.
The P-value
is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.
My computations from R statistical software are shown below:
(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565
Also, Minitab statistical software gives the following relevant output:
One-Sample Z
Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2
N Mean SE Mean 99% CI Z P
4 7.540 0.100 (7.282, 7.798) 0.40 0.689
Because your hypothetical mean $mu = 7.5$ is included in the 99%
confidence interval, you would not reject at the 1% level.
answered Jan 27 at 10:42
BruceETBruceET
36k71540
$endgroup$
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
add a comment |
$begingroup$
You have written the population variance as $sigma^2 = 0.004,$ not $0.04.$
Also, in the problem you state that $bar X = 7.54,$ but in your computation you have $7.48.$
Otherwise, you seem to be on the right track.
After fixing your typos, you should have $Z = 0.4.$ Then at level $alpha = 0.01 = 1%,$
you would reject if $|Z| > 2.576,$ so you cannot reject.
The P-value
is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.
My computations from R statistical software are shown below:
(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565
Also, Minitab statistical software gives the following relevant output:
One-Sample Z
Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2
N Mean SE Mean 99% CI Z P
4 7.540 0.100 (7.282, 7.798) 0.40 0.689
Because your hypothetical mean $mu = 7.5$ is included in the 99%
confidence interval, you would not reject at the 1% level.
answered Jan 27 at 10:42
BruceETBruceET
36k71540
$endgroup$
You have written the population variance as $sigma^2 = 0.004,$ not $0.04.$
Also, in the problem you state that $bar X = 7.54,$ but in your computation you have $7.48.$
Otherwise, you seem to be on the right track.
After fixing your typos, you should have $Z = 0.4.$ Then at level $alpha = 0.01 = 1%,$
you would reject if $|Z| > 2.576,$ so you cannot reject.
The P-value
is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.
My computations from R statistical software are shown below:
(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565
Also, Minitab statistical software gives the following relevant output:
One-Sample Z
Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2
N Mean SE Mean 99% CI Z P
4 7.540 0.100 (7.282, 7.798) 0.40 0.689
Because your hypothetical mean $mu = 7.5$ is included in the 99%
confidence interval, you would not reject at the 1% level.
answered Jan 27 at 10:42
BruceETBruceET
36k71540
edited Jan 27 at 11:07
answered Jan 27 at 10:42
BruceETBruceET
36k71540
answered Jan 27 at 10:42
BruceETBruceET
36k71540
answered Jan 27 at 10:42
BruceETBruceET
36k71540
36k71540
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
add a comment |
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
1
1
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
$begingroup$
Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$?
$endgroup$
– Mark Jacon
Jan 27 at 11:38
1
1
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
$begingroup$
Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $sigma$ is known is of form $bar X pm 2.576sigma/sqrt{n}.$ Try it and compare with CI in Minitab output.
$endgroup$
– BruceET
Jan 27 at 19:29
add a comment |
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