Question

In a study of the relationship of the shape of a tablet to its dissolution time, 6 disk-shaped ibuprofen tablets and 8 oval-shaped ibuprofen tablets were dissolved in water. The dissolve times, in seconds, were as follows:
Disk: 269.0 249.3 255.2 252.7 247.0 261.6
Oval: 268.8 260.0 273.5 253.9 278.5 289.4 261.6 280.2
Can you conclude that the mean dissolve times differ between the two shapes?

Answer 1

Answer:

$t=\frac{255.8-270.74}{\sqrt{\frac{(8.215)^2}{6}+\frac{(11.903)^2}{8}}}}=-2.788$  

$p_v =2*P(t_{(12)}<-2.788)=0.0164$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that we have significant difference in the means of the two groups

Step-by-step explanation:

Data given and notation

Disk:[269.0 249.3 255.2 252.7 247.0 261.6]

Oval:[ 268.8 260.0 273.5 253.9 278.5 289.4 261.6 280.2]

$\bar X_{D}=255.8$ represent the mean for the sample Disk

$\bar X_{O}=270.74$ represent the mean for the sample Oval

$s_{D}=8.215$ represent the sample standard deviation for the sample Disk

$s_{O}=11.903$ represent the sample standard deviation for the sample Oval

$n_{D}=6$ sample size for the group Disk

$n_{O}=8$ sample size for the group Oval

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means are different, the system of hypothesis would be:

Null hypothesis:$\mu_{D} = \mu_{O}$

Alternative hypothesis:$\mu_{D} \neq \mu_{O}$

If we analyze the size for the samples both are less than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{D}-\bar X_{O}}{\sqrt{\frac{s^2_{D}}{n_{D}}+\frac{s^2_{O}}{n_{O}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{255.8-270.74}{\sqrt{\frac{(8.215)^2}{6}+\frac{(11.903)^2}{8}}}}=-2.788$  

Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

$df=n_{D}+n_{O}-2=6+8-2=12$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(12)}<-2.788)=0.0164$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that we have significant difference in the means of the two groups