Question on a hypothesis test The 2019 Stack Overflow Developer Survey Results Are InPlease verify this Proportion Hypothesis Test (Statistics)Hypothesis test questionHypothesis testing: Test statistic, P-value and significance levelsfind distribution of hypothesis testing?find the value of the test statisticFind the value for k that gives a level of significance of 0.05, when given a null hypothesis that is to be rejected when $y_1*y_2 leq k$Statistical test with proportions and infinite population.Computing Type II Error for a One-Sided Normal Test.One-tailed hypothesis decision based on a two-tailed Z testHypothesis testing variance using sample mean
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Question on a hypothesis test
The 2019 Stack Overflow Developer Survey Results Are InPlease verify this Proportion Hypothesis Test (Statistics)Hypothesis test questionHypothesis testing: Test statistic, P-value and significance levelsfind distribution of hypothesis testing?find the value of the test statisticFind the value for k that gives a level of significance of 0.05, when given a null hypothesis that is to be rejected when $y_1*y_2 leq k$Statistical test with proportions and infinite population.Computing Type II Error for a One-Sided Normal Test.One-tailed hypothesis decision based on a two-tailed Z testHypothesis testing variance using sample mean
$begingroup$
So I have population mean (for average production on some farms) = $3000$, and a test for a new fertilizer has sample size $n = 70$, sample mean = $3120$, sample standard deviation = $578$.
The company making the fertilizer claims that it knows that the fertilizer improves the production. Perform the appropriate test with a significance level of $0.05$.
So I computed the test statistic
$$frac3120 - 3000frac578sqrt70 = 1.73 ,$$
and since this is a one-sided test, we will compare with the respective normal quantile for 0.95 $= 1.96$.
Now is the correct null hypothesis $H_0: μ>3000$ or $H_0:μ leq 3000$ (versus their respective negations)? And, whichever the case is, which is the correct rejection region?
probability statistics
edited Mar 23 at 1:43
Ertxiem
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
$endgroup$
|
show 2 more comments
$begingroup$
So I have population mean (for average production on some farms) = $3000$, and a test for a new fertilizer has sample size $n = 70$, sample mean = $3120$, sample standard deviation = $578$.
The company making the fertilizer claims that it knows that the fertilizer improves the production. Perform the appropriate test with a significance level of $0.05$.
So I computed the test statistic
$$frac3120 - 3000frac578sqrt70 = 1.73 ,$$
and since this is a one-sided test, we will compare with the respective normal quantile for 0.95 $= 1.96$.
Now is the correct null hypothesis $H_0: μ>3000$ or $H_0:μ leq 3000$ (versus their respective negations)? And, whichever the case is, which is the correct rejection region?
probability statistics
edited Mar 23 at 1:43
Ertxiem
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
$endgroup$
1
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
$endgroup$
– littleO
Mar 23 at 0:45
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
$endgroup$
– JBuck
Mar 23 at 1:05
1
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
$endgroup$
– Ertxiem
Mar 23 at 1:14
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
$endgroup$
– JBuck
Mar 23 at 1:19
1
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
$endgroup$
– Ertxiem
Mar 23 at 1:22
|
show 2 more comments
$begingroup$
So I have population mean (for average production on some farms) = $3000$, and a test for a new fertilizer has sample size $n = 70$, sample mean = $3120$, sample standard deviation = $578$.
The company making the fertilizer claims that it knows that the fertilizer improves the production. Perform the appropriate test with a significance level of $0.05$.
So I computed the test statistic
$$frac3120 - 3000frac578sqrt70 = 1.73 ,$$
and since this is a one-sided test, we will compare with the respective normal quantile for 0.95 $= 1.96$.
Now is the correct null hypothesis $H_0: μ>3000$ or $H_0:μ leq 3000$ (versus their respective negations)? And, whichever the case is, which is the correct rejection region?
probability statistics
edited Mar 23 at 1:43
Ertxiem
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
$endgroup$
So I have population mean (for average production on some farms) = $3000$, and a test for a new fertilizer has sample size $n = 70$, sample mean = $3120$, sample standard deviation = $578$.
The company making the fertilizer claims that it knows that the fertilizer improves the production. Perform the appropriate test with a significance level of $0.05$.
So I computed the test statistic
$$frac3120 - 3000frac578sqrt70 = 1.73 ,$$
and since this is a one-sided test, we will compare with the respective normal quantile for 0.95 $= 1.96$.
Now is the correct null hypothesis $H_0: μ>3000$ or $H_0:μ leq 3000$ (versus their respective negations)? And, whichever the case is, which is the correct rejection region?
probability statistics
probability statistics
edited Mar 23 at 1:43
Ertxiem
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
edited Mar 23 at 1:43
Ertxiem
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
edited Mar 23 at 1:43
Ertxiem
671212
edited Mar 23 at 1:43
Ertxiem
671212
edited Mar 23 at 1:43
Ertxiem
671212
671212
asked Mar 23 at 0:40
JBuckJBuck
1218
asked Mar 23 at 0:40
JBuckJBuck
1218
asked Mar 23 at 0:40
JBuckJBuck
1218
1218
1
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
$endgroup$
– littleO
Mar 23 at 0:45
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
$endgroup$
– JBuck
Mar 23 at 1:05
1
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
$endgroup$
– Ertxiem
Mar 23 at 1:14
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
$endgroup$
– JBuck
Mar 23 at 1:19
1
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
$endgroup$
– Ertxiem
Mar 23 at 1:22
|
show 2 more comments
1
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
$endgroup$
– littleO
Mar 23 at 0:45
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
$endgroup$
– JBuck
Mar 23 at 1:05
1
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
$endgroup$
– Ertxiem
Mar 23 at 1:14
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
$endgroup$
– JBuck
Mar 23 at 1:19
1
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
$endgroup$
– Ertxiem
Mar 23 at 1:22
1
1
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
$endgroup$
– littleO
Mar 23 at 0:45
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
$endgroup$
– littleO
Mar 23 at 0:45
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
$endgroup$
– JBuck
Mar 23 at 1:05
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
$endgroup$
– JBuck
Mar 23 at 1:05
1
1
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
$endgroup$
– Ertxiem
Mar 23 at 1:14
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
$endgroup$
– Ertxiem
Mar 23 at 1:14
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
$endgroup$
– JBuck
Mar 23 at 1:19
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
$endgroup$
– JBuck
Mar 23 at 1:19
1
1
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
$endgroup$
– Ertxiem
Mar 23 at 1:22
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
$endgroup$
– Ertxiem
Mar 23 at 1:22
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The population standard deviation is not known, so looking ahead, you should understand that this situation really should be analyzed using a t test. However, for for such a large sample size as $n = 70,$ you can get by assuming that the population SD is $sigma=578.$
T test--Population SD unknown: Here is a printout of the exact t test from Minitab (slightly edited for relevance).
One-Sample T
Test of μ = 3000 vs > 3000
N Mean StDev SE Mean T P
70 3120.0 578.0 69.1 1.74 0.043
(a) From printed t-tables, the 5% critical value for this one-sided t test is $c = 1.667;$
because $T = 1.74 > 1.667,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.43 is less than 5%.
Z test--Population SD assumed: For comparison, this is a printout for your approximate z test. Here you have to "lie to the software," entering the sample standard $S = 578$ as if it were the population SD $sigma.$ The (slightly edited) output is as follows:
One-Sample Z
Test of μ = 3000 vs > 3000
The assumed population standard deviation = 578
N Mean SE Mean Z P
70 3120.0 69.1 1.74 0.041
(a) From printed normal tables, the 5% critical value for this one-sided z test is $c = 1.645;$
because $T = 1.74 > 1.645,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.41 is less than 5%.
Summary: Either way, the observed mean $bar X = 3120$ is enough larger than
the hypothetical population mean $mu = 3000$ to reject the null hypothesis $H_0: mu = 3000$ against the one-sided alternative $H_a: mu > 3000$.
However, if the "z statistic" had turned out to be $Z = 1.650$ (between 1.667 and 1.645), then the "z test" would reject $H_0$ (just barely) and the t test would not. So if you're testing exactly at a particular significance level, the approximate z test is not always a good substitute for the t test--even with s sample size as large as 70.
Also, there can be a big difference between a one-sided test and a two-sided test: For your data, a test against a two-sided alternative $H_a: mu ne 3000$ would
have a P-value of about 8% and you could not reject at the 5% level.
[By doing a one-sided test, you are, in effect, ignoring the possibility that
the new fertilizer might be worse.]
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
$endgroup$
add a comment |
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$begingroup$
The population standard deviation is not known, so looking ahead, you should understand that this situation really should be analyzed using a t test. However, for for such a large sample size as $n = 70,$ you can get by assuming that the population SD is $sigma=578.$
T test--Population SD unknown: Here is a printout of the exact t test from Minitab (slightly edited for relevance).
One-Sample T
Test of μ = 3000 vs > 3000
N Mean StDev SE Mean T P
70 3120.0 578.0 69.1 1.74 0.043
(a) From printed t-tables, the 5% critical value for this one-sided t test is $c = 1.667;$
because $T = 1.74 > 1.667,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.43 is less than 5%.
Z test--Population SD assumed: For comparison, this is a printout for your approximate z test. Here you have to "lie to the software," entering the sample standard $S = 578$ as if it were the population SD $sigma.$ The (slightly edited) output is as follows:
One-Sample Z
Test of μ = 3000 vs > 3000
The assumed population standard deviation = 578
N Mean SE Mean Z P
70 3120.0 69.1 1.74 0.041
(a) From printed normal tables, the 5% critical value for this one-sided z test is $c = 1.645;$
because $T = 1.74 > 1.645,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.41 is less than 5%.
Summary: Either way, the observed mean $bar X = 3120$ is enough larger than
the hypothetical population mean $mu = 3000$ to reject the null hypothesis $H_0: mu = 3000$ against the one-sided alternative $H_a: mu > 3000$.
However, if the "z statistic" had turned out to be $Z = 1.650$ (between 1.667 and 1.645), then the "z test" would reject $H_0$ (just barely) and the t test would not. So if you're testing exactly at a particular significance level, the approximate z test is not always a good substitute for the t test--even with s sample size as large as 70.
Also, there can be a big difference between a one-sided test and a two-sided test: For your data, a test against a two-sided alternative $H_a: mu ne 3000$ would
have a P-value of about 8% and you could not reject at the 5% level.
[By doing a one-sided test, you are, in effect, ignoring the possibility that
the new fertilizer might be worse.]
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
$endgroup$
add a comment |
$begingroup$
The population standard deviation is not known, so looking ahead, you should understand that this situation really should be analyzed using a t test. However, for for such a large sample size as $n = 70,$ you can get by assuming that the population SD is $sigma=578.$
T test--Population SD unknown: Here is a printout of the exact t test from Minitab (slightly edited for relevance).
One-Sample T
Test of μ = 3000 vs > 3000
N Mean StDev SE Mean T P
70 3120.0 578.0 69.1 1.74 0.043
(a) From printed t-tables, the 5% critical value for this one-sided t test is $c = 1.667;$
because $T = 1.74 > 1.667,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.43 is less than 5%.
Z test--Population SD assumed: For comparison, this is a printout for your approximate z test. Here you have to "lie to the software," entering the sample standard $S = 578$ as if it were the population SD $sigma.$ The (slightly edited) output is as follows:
One-Sample Z
Test of μ = 3000 vs > 3000
The assumed population standard deviation = 578
N Mean SE Mean Z P
70 3120.0 69.1 1.74 0.041
(a) From printed normal tables, the 5% critical value for this one-sided z test is $c = 1.645;$
because $T = 1.74 > 1.645,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.41 is less than 5%.
Summary: Either way, the observed mean $bar X = 3120$ is enough larger than
the hypothetical population mean $mu = 3000$ to reject the null hypothesis $H_0: mu = 3000$ against the one-sided alternative $H_a: mu > 3000$.
However, if the "z statistic" had turned out to be $Z = 1.650$ (between 1.667 and 1.645), then the "z test" would reject $H_0$ (just barely) and the t test would not. So if you're testing exactly at a particular significance level, the approximate z test is not always a good substitute for the t test--even with s sample size as large as 70.
Also, there can be a big difference between a one-sided test and a two-sided test: For your data, a test against a two-sided alternative $H_a: mu ne 3000$ would
have a P-value of about 8% and you could not reject at the 5% level.
[By doing a one-sided test, you are, in effect, ignoring the possibility that
the new fertilizer might be worse.]
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
$endgroup$
add a comment |
$begingroup$
The population standard deviation is not known, so looking ahead, you should understand that this situation really should be analyzed using a t test. However, for for such a large sample size as $n = 70,$ you can get by assuming that the population SD is $sigma=578.$
T test--Population SD unknown: Here is a printout of the exact t test from Minitab (slightly edited for relevance).
One-Sample T
Test of μ = 3000 vs > 3000
N Mean StDev SE Mean T P
70 3120.0 578.0 69.1 1.74 0.043
(a) From printed t-tables, the 5% critical value for this one-sided t test is $c = 1.667;$
because $T = 1.74 > 1.667,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.43 is less than 5%.
Z test--Population SD assumed: For comparison, this is a printout for your approximate z test. Here you have to "lie to the software," entering the sample standard $S = 578$ as if it were the population SD $sigma.$ The (slightly edited) output is as follows:
One-Sample Z
Test of μ = 3000 vs > 3000
The assumed population standard deviation = 578
N Mean SE Mean Z P
70 3120.0 69.1 1.74 0.041
(a) From printed normal tables, the 5% critical value for this one-sided z test is $c = 1.645;$
because $T = 1.74 > 1.645,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.41 is less than 5%.
Summary: Either way, the observed mean $bar X = 3120$ is enough larger than
the hypothetical population mean $mu = 3000$ to reject the null hypothesis $H_0: mu = 3000$ against the one-sided alternative $H_a: mu > 3000$.
However, if the "z statistic" had turned out to be $Z = 1.650$ (between 1.667 and 1.645), then the "z test" would reject $H_0$ (just barely) and the t test would not. So if you're testing exactly at a particular significance level, the approximate z test is not always a good substitute for the t test--even with s sample size as large as 70.
Also, there can be a big difference between a one-sided test and a two-sided test: For your data, a test against a two-sided alternative $H_a: mu ne 3000$ would
have a P-value of about 8% and you could not reject at the 5% level.
[By doing a one-sided test, you are, in effect, ignoring the possibility that
the new fertilizer might be worse.]
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
$endgroup$
The population standard deviation is not known, so looking ahead, you should understand that this situation really should be analyzed using a t test. However, for for such a large sample size as $n = 70,$ you can get by assuming that the population SD is $sigma=578.$
T test--Population SD unknown: Here is a printout of the exact t test from Minitab (slightly edited for relevance).
One-Sample T
Test of μ = 3000 vs > 3000
N Mean StDev SE Mean T P
70 3120.0 578.0 69.1 1.74 0.043
(a) From printed t-tables, the 5% critical value for this one-sided t test is $c = 1.667;$
because $T = 1.74 > 1.667,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.43 is less than 5%.
Z test--Population SD assumed: For comparison, this is a printout for your approximate z test. Here you have to "lie to the software," entering the sample standard $S = 578$ as if it were the population SD $sigma.$ The (slightly edited) output is as follows:
One-Sample Z
Test of μ = 3000 vs > 3000
The assumed population standard deviation = 578
N Mean SE Mean Z P
70 3120.0 69.1 1.74 0.041
(a) From printed normal tables, the 5% critical value for this one-sided z test is $c = 1.645;$
because $T = 1.74 > 1.645,$ you can reject $H_0: mu = 5000.$
(b) From computer output, you can reject $H_0$ because the P-value 0.41 is less than 5%.
Summary: Either way, the observed mean $bar X = 3120$ is enough larger than
the hypothetical population mean $mu = 3000$ to reject the null hypothesis $H_0: mu = 3000$ against the one-sided alternative $H_a: mu > 3000$.
However, if the "z statistic" had turned out to be $Z = 1.650$ (between 1.667 and 1.645), then the "z test" would reject $H_0$ (just barely) and the t test would not. So if you're testing exactly at a particular significance level, the approximate z test is not always a good substitute for the t test--even with s sample size as large as 70.
Also, there can be a big difference between a one-sided test and a two-sided test: For your data, a test against a two-sided alternative $H_a: mu ne 3000$ would
have a P-value of about 8% and you could not reject at the 5% level.
[By doing a one-sided test, you are, in effect, ignoring the possibility that
the new fertilizer might be worse.]
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
edited Mar 23 at 20:09
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
answered Mar 23 at 17:38
BruceETBruceET
36.3k71540
36.3k71540
add a comment |
add a comment |
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1
$begingroup$
No, the null hypothesis should be the more conservative position that the new fertilizer does not improve production.
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– littleO
Mar 23 at 0:45
$begingroup$
@littleO I see, so the rejection region is $z > 1.96$ and therefore we cannot reject $H_0 : μ=<3000$ ?
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– JBuck
Mar 23 at 1:05
1
$begingroup$
If it is an unilateral test you should not use the bilateral value of $1.96$. Furthermore, if you have the sample standard deviation (and not the population), you should use t-Student.
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– Ertxiem
Mar 23 at 1:14
$begingroup$
@Ertxiem As for t-student, you might be correct, but we haven't learned it yet, so the exercise is meant to be solved along these lines. Second, what value should I use since the test is unilateral?
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– JBuck
Mar 23 at 1:19
1
$begingroup$
I'm assuming that you have a table for the Normal distribution. The value of $1.96$ corresponds to 2.5% of the $z$ values above it (and 2.5% of the $z$ values below $-1.96$, for a total of 5%). If you look into the table, you should find a value of $z$ that has 5% of the values above it.
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– Ertxiem
Mar 23 at 1:22