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Step-by-step explanation:
Hello!
The objective of this experiment is to test wich paint formulation reduces the drying time of primer paint.
The formulation 1 has the standard chemistry and the formulation 2 has a new drying ingredient, that hypothetically reduces drying time.
You have two study variables:
X₁: Drying time of the primer paint formulated with the standard chemistry.
X₂: Drying time of the primer paint formulated with the addition of the new drying ingredient.
It is known that the standard deviation of drying time is 8 minutes and this variability should not be unaffected by the addition of the new ingredient, in other words, the standard deviation for both variables is the same. δ₁=δ₂= 8 min
Two samples were taken and the drying time for both formulations was determined.
Sample 1
n₁= 10
X[bar]₁= 121 min
Sample 2
n₂= 10
X[bar]₂= 112 min
If the new ingredient reduces the drying time, we would expect that the average drying time of the formulation 2 will be less than the average time of the formulation 1, symbolically: μ₁ < μ₂
The hypotheses are:
H₀: μ₁ ≥ μ₂
H₁: μ₁ < μ₂
α: 0.05
Assuming that both variables have a normal distribution, the statistic to use is:
[tex]Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sigma\sqrt{\frac{1}{n_1} + \frac{1}{n_2} } }[/tex]
[tex]Z_{H_0}= \frac{(121-112)-0}{8* \sqrt{\frac{1}{10}+\frac{1}{10} } } = 2.516[/tex]
The critical region of this test is one-tailed (left) and so is the p-value of the test. You can calculate the p-value manually or use statistical software to obtain it.
The p-value is: 0.0059
The decision rule using the p-value is:
If p-value ≤ α, then you reject the null hypothesis.
If p-value > α, then you do not reject the null hypothesis.
Since the p-value is greater than α, then the decision is to reject the null hypothesis.
With a level of significance of 5%, there is significant evidence to conclude that the new ingredient reduces the drying time of the primer paint.
I hope it helps!
Using the z-distribution, it is found that since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to conclude that the time has been reduced, that is, the new ingredient is efficient.
At the null hypothesis, it is tested if there is no reduction in time, that is, the subtraction of the means is at least 0.
[tex]H_0: \mu_2 - \mu_1 \geq 0[/tex]
At the alternative hypothesis, it is tested if there is reduction, that is:
[tex]H_1: \mu_2 - \mu_1 < 0[/tex]
We have the standard deviation for the population, hence, the z-distribution is used.
The standard errors are:
[tex]s_1 = \frac{8}{\sqrt{10}} = 2.5298[/tex]
[tex]s_2 = \frac{8}{\sqrt{20}} = 1.78885[/tex]
For the distribution of differences, the mean and the standard error are given by:
[tex]\overline{x} = \mu_2 - \mu_1 = 112 - 121 = -9[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{2.5298^2 + 1.78885^2} = 3.0984[/tex]
The test statistic is given by:
[tex]z = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Hence:
[tex]z = \frac{\overline{x} - \mu}{s}[/tex]
[tex]z = \frac{-9 - 0}{3.0984}[/tex]
[tex]z = -2.9[/tex]
The critical value for a left-tailed test, as we are testing if the mean is less than a value, using the z-distribution with a significance level of 0.05, is of [tex]z^{\ast} = -1.645[/tex].
Since the test statistic is less than the critical value for the left-tailed test, it is found that there is enough evidence to conclude that the time has been reduced, that is, the new ingredient is efficient.
To learn more about the use of the z-distribution to test an hypothesis, you can take a look at https://brainly.com/question/25676691
