# Is the new fertilizer more efficient than the old one?

#### mathmari

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Hey!! In an effort to evaluate the quality of a new fertilizer we selected 12 fields which we split exactly in the middle. In the one half we used the old fertilizer and in the other half we used the new fertilizer. Then we measured in tonnes the crop that was harvested from each piece of land.

The results are in the following table: 1. Should we use the dependent or the independent samples method to compare crops from the old and the new fertilizer?
2. At significance level $a = 10\%$ can we say that the new fertilizer is more efficient than the old one?
3. If we used the method of independent samples, what would we conclude? Compare the tests of questions 2 and 3.

I have done the following:

1. The two samples are dependent since they come from the same fields. The fields with the most crop with the old fertilizer will probably have the most crop also with the new fertilizer.

Is that correct? 2. Do we calculate the mean value and define the null hypothesis and the alternativ hypothesis? Or how can we check that? #### Klaas van Aarsen

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Hey!! In an effort to evaluate the quality of a new fertilizer we selected 12 fields which we split exactly in the middle. In the one half we used the old fertilizer and in the other half we used the new fertilizer. Then we measured in tonnes the crop that was harvested from each piece of land.

The results are in the following table:

1. Should we use the dependent or the independent samples method to compare crops from the old and the new fertilizer?
2. At significance level $a = 10\%$ can we say that the new fertilizer is more efficient than the old one?
3. If we used the method of independent samples, what would we conclude? Compare the tests of questions 2 and 3.

I have done the following:

[*] The two samples are dependent since they come from the same fields. The fields with the most crop with the old fertilizer will probably have the most crop also with the new fertilizer.

Is that correct? Hey mathmari !!

Yep.
More specifically a paired dependent method. [*] Do we calculate the mean value and define the null hypothesis and the alternativ hypothesis? Or how can we check that? Proper order is to define the hypotheses first. Then select a method. Measure a sample. And only then can we calculate a mean value. #### mathmari

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Proper order is to define the hypotheses first. Then select a method. Measure a sample. And only then can we calculate a mean value. The null hypothesis is that the new fertilizer is more efficient than the old one and therefore the alternative one is that the new fertilizer isn't more efficient than the old one, or not? If that is correct,how do we continue? How do we measure a sample? Do we choose by ourselves a sample size? #### Klaas van Aarsen

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The null hypothesis is that the new fertilizer is more efficient than the old one and therefore the alternative one is that the new fertilizer isn't more efficient than the old one, or not?
I don't think so. The null hypothesis is usually that there is no difference. And the alternative hypothesis is that there is a difference - possibly in a certain direction.

Moreover, the problem statement does not mention anything about whether the new fertilizer would be more efficient or not.
Instead it only says to 'evaluate the new fertilizer'. If that is correct,how do we continue? How do we measure a sample? Do we choose by ourselves a sample size?
To be fair, the sample has already been measured. It is given to us as part of the problem statement.
It's just that when we define the hypotheses, we must 'pretend' it wasn't measured yet, as it is a step that is supposed to come before. #### mathmari

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The null hypothesis is usually that there is no difference. And the alternative hypothesis is that there is a difference - possibly in a certain direction.

Moreover, the problem statement does not mention anything about whether the new fertilizer would be more efficient or not.
Instead it only says to 'evaluate the new fertilizer'. So do you mean that the null hypothesis assumes that both fertilizers have the same results and the alternative hypothesis assumes that the new fertilizer is better, i.e. gives more crop? #### Klaas van Aarsen

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So do you mean that the null hypothesis assumes that both fertilizers have the same results
Yes. and the alternative hypothesis assumes that the new fertilizer is better, i.e. gives more crop?
Since the problem statement does not suggest that the new fertilizer is supposed to be 'better' nor 'more effective' nor 'has greater yield' nor anything of that sort, I believe the alternative hypothesis should simply be that the new fertilizer is different from the old fertilizer. #### mathmari

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Yes. Since the problem statement does not suggest that the new fertilizer is supposed to be 'better' nor 'more effective' nor 'has greater yield' nor anything of that sort, I believe the alternative hypothesis should simply be that the new fertilizer is different from the old fertilizer. Ahh ok!

So we have the following:

The null hypothesis assumes that both fertilizers have the same results and the alternative hypothesis assumes that the new fertilizer is different from the old fertilizer.

Does this mean that at the null hypothesis the mean values for the old and the new fertilizer respectively are the same and at the alternative they are different? Or do we not consider the mean values? #### Klaas van Aarsen

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Ahh ok!

So we have the following:

The null hypothesis assumes that both fertilizers have the same results and the alternative hypothesis assumes that the new fertilizer is different from the old fertilizer.

Does this mean that at the null hypothesis the mean values for the old and the new fertilizer respectively are the same and at the alternative they are different? Or do we not consider the mean values?
We will look at the mean values and decide whether they are sufficiently different to consider them significantly different. That is, if we can conclude that the probability that we are wrong in concluding they are different is less than $\alpha=10\%$.
It may still be that the means are close enough together that we cannot be sufficiently sure they are different. #### mathmari

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We will look at the mean values and decide whether they are sufficiently different to consider them significantly different. That is, if we can conclude that the probability that we are wrong in concluding they are different is less than $\alpha=10\%$.
It may still be that the means are close enough together that we cannot be sufficiently sure they are different. Ok.. So we have stated the null and the alternative hypothesis.

Do we apply now a t-test? Or what do we do next? #### Klaas van Aarsen

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Ok.. So we have stated the null and the alternative hypothesis.

Do we apply now a t-test? Or what do we do next?
Yep. A t-test is appropriate here. After all, a t-test is used to test if population means are different.

#### mathmari

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Yep. A t-test is appropriate here. After all, a t-test is used to test if population means are different.
We have the following: So since t-exp > t-crit is not true, we accept the null hypothesis.

Is everything correct? If yes, could you give me a hint for the question 3? Do we consider that the two fertilizers are not used in the same fields orhow do we consider the method of independent samples? #### Klaas van Aarsen

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We have the following:

So since t-exp > t-crit is not true, we accept the null hypothesis.

Is everything correct?
t-tests come in different variants, each with their own formulas.
Notably:
• the 1-sample t-test,
• the independent 2-sample t-test,
• the dependent 2-sample t-test with unequal variances,
• the dependent 2-sample t-test with equal variances,
• the dependent paired t-test.
Since we have dependent pairs of measurements, shouldn't we do a dependent paired t-test? #### mathmari

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t-tests come in different variants, each with their own formulas.
Notably:
• the 1-sample t-test,
• the independent 2-sample t-test,
• the dependent 2-sample t-test with unequal variances,
• the dependent 2-sample t-test with equal variances,
• the dependent paired t-test.
Since we have dependent pairs of measurements, shouldn't we do a dependent paired t-test? Ah ok! I used a command of Excel and I got the following: Are these the values that we need? #### Klaas van Aarsen

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Ah ok! I used a command of Excel and I got the following:

Are these the values that we need? I am not sure. Your values look as if you took a mean of each sample, but that would be wrong.
The proper evaluation is to consider a single sample that consists of the differences of each 'pair'.
Consequently we have a single sample of differences.
And we would compare its mean to '0', which is what we would have if 'there is no difference'. #### mathmari

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I am not sure. Your values look as if you took a mean of each sample, but that would be wrong.
The proper evaluation is to consider a single sample that consists of the differences of each 'pair'.
Consequently we have a single sample of differences.
And we would compare its mean to '0', which is what we would have if 'there is no difference'. There is no command in Excel for that, is there? We have to do that by ourselves, or not? #### Klaas van Aarsen

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There is no command in Excel for that, is there? We have to do that by ourselves, or not?
The 'trick' if there is such a thing, is to calculate each difference and treat the results as a single sample that we compare to zero. #### mathmari

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The 'trick' if there is such a thing, is to calculate each difference and treat the results as a single sample that we compare to zero. We have the following table: We have the hypotheses: $H_0: \ \mu_d=0$ and $H_0: \ \mu_d\neq 0$.

We have that $\overline{d}=\frac{24}{12}=2$, $s_d=\sqrt{\frac{170}{12}}=3.76$.

The test statistic is \begin{equation*}t=\frac{\overline{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}=\frac{2-0}{\frac{3.76}{\sqrt{12}}}=\frac{2-0}{\frac{3.76}{\sqrt{12}}}=1.84261\end{equation*}

For the degree of freedom, since we are testing $12$ values of difference, we have $df=12-1=11$.

Is everything correct so far? Now it is left to calculate the p-value, or not? #### Klaas van Aarsen

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We have the following table:

We have the hypotheses: $H_0: \ \mu_d=0$ and $H_0: \ \mu_d\neq 0$.
I have reread the question, which says:
2. At significance level a=10% can we say that the new fertilizer is more efficient than the old one?
It means that the alternative hypothesis should be $H_1:\,\mu_d > 0$. We have that $\overline{d}=\frac{24}{12}=2$, $s_d=\sqrt{\frac{170}{12}}=3.76$.
When we calculate a sample standard deviation, shouldn't we divide by $n-1=12-1$? The test statistic is \begin{equation*}t=\frac{\overline{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}=\frac{2-0}{\frac{3.76}{\sqrt{12}}}=\frac{2-0}{\frac{3.76}{\sqrt{12}}}=1.84261\end{equation*}

For the degree of freedom, since we are testing $12$ values of difference, we have $df=12-1=11$.

Is everything correct so far? Now it is left to calculate the p-value, or not?
After correction of $s_d$, the next step is indeed to calculate the p-value. #### mathmari

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I have reread the question, which says:

It means that the alternative hypothesis should be $H_1:\,\mu_d > 0$. When we calculate a sample standard deviation, shouldn't we divide by $n-1=12-1$? After correction of $s_d$, the next step is indeed to calculate the p-value. Ahh ok! So we have the following:

We have the hypotheses: $H_0: \ \mu_d=0$ and $H_1: \ \mu_d> 0$.

We have that $\overline{d}=\frac{24}{12}=2$, $s_d=\sqrt{\frac{170}{12-1}}=\sqrt{\frac{170}{11}}=3.93$.

The test statistic is \begin{equation*}t=\frac{\overline{d}-\mu_d}{\frac{s_d}{\sqrt{n}}}=\frac{2-0}{\frac{3.93}{\sqrt{12}}}=\frac{2}{\frac{3.93}{\sqrt{12}}}=1.76290\end{equation*}

For the degree of freedom, since we are testing $12$ values of difference, we have $df=12-1=11$.

Using the R-program command "pt(1.76290,df=11, lower.tail=TRUE )" we get the p-value $0.9471801$.

We have that $\text{p-value} > \alpha$ and so we do not reject $H_0$, there is not enough evidence to show that the new fertilizer is more efficient than the old one.

Is everything correct? #### Klaas van Aarsen

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That p-value does not look correct. A t-value of 1.76 should be more or less significant. I think we have the wrong formula in R. #### mathmari

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That p-value does not look correct. A t-value of 1.76 should be more or less sofnificant. I think we have the wrong formula in R. Ok. Now I tried it with Excel, I used the command [M]TDIST(ABS(1,7629);11;1)[/M] and I get $0.052819887$. Now we have that $\text{p-value} < \alpha$ and so we accept the null hypothesis.

Is that correct? That would mean that the new fertilizer has the same results as the old one, or not? #### Klaas van Aarsen

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Ok. Now I tried it with Excel, I used the command [M]TDIST(ABS(1,7629);11;1)[/M] and I get $0.052819887$. Now we have that $\text{p-value} < \alpha$ and so we accept the null hypothesis.

Is that correct? That would mean that the new fertilizer has the same results as the old one, or not? Nope. A high t-value implies that the means are far apart, doesn't it? What does it mean again that $p<\alpha$? #### mathmari

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Nope. A high t-value implies that the means are far apart, doesn't it? What does it mean again that $p<\alpha$? Oh yes, you're right. I confused that. So since $p<\alpha$ we reject the null hypothesis and so the two fertilizer don't have the same results, and so the new one is more efficient than the old one, right? #### Klaas van Aarsen

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Oh yes, you're right. I confused that. So since $p<\alpha$ we reject the null hypothesis and so the two fertilizer don't have the same results, and so the new one is more efficient than the old one, right? Yep.
We have a confidence of more than 90% that the new fertilizer is more efficient than the old one. #### mathmari

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Yep.
We have a confidence of more than 90% that the new fertilizer is more efficient than the old one. Great!! Could you give me a hint for the question 3? Do we consider that the two fertilizers are not used in the same fields or how do we consider the method of independent samples? 