Question

In a study of the relationships 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 time is less for disk shaped tablets than for mean dissolve time for oval shaped tablets? Assume that the two samples come from normal distributions and σdisk= σoval.

a. Carry out the appropriate test at the 5% level. Be sure to show the hypothesis statements.
b. Generate the appropriate 95% one-sided confidence interval.

Answer 1

Answer:

Step-by-step explanation:

Hello!

a.

The objective is to study the relationship between the shape of an ibuprofen tablet and its dissolution time.

For these two independent samples of tablets from different shapes where taken and their dissolution times measured:

Sample 1: Disk.shaped tablets

n₁= 6

X[bar]₁= 255.8

S₁= 8.22

Sample 2: Oval-shaped tablets

n₂=8

X[bar]₂= 270.74

S₂= 11.90

Assuming that the population variances are equal and both samples come from normal distributions you need to test if the average dissolution time of the disk-shaped tablets is less than the average dissolution time of the oval-shaped tablets, symbolically:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

Considering the given information about both populations, the statistic to use for this test is a Student t for independent samples with pooled sample variance:

$t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

$Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}= \frac{5*(8.22)^2+7*(11.9)^2}{6*8-2}$

Sa²= 110.76

Sa= 10.52

$t_{H_0}= \frac{(255.8-270.74)-0}{10.52\sqrt{\frac{1}{6} +\frac{1}{8} } } = -2.629= -2.63$

This test is one-tailed to the left, meaning that you will reject the null hypothesis to small values of t, the p-value has the same direction and you can calculate it as:

P(t₁₂≤-2.63)= 0.0110

Since the p-value= 0.0110 is less than the significance level α: 0.05, the decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average dissolution time of the disk-shaped ibuprofen tablets is less than the average dissolution time of the oval-shaped ibuprofen tablets.

b.

(X[bar]₁-X[bar]₂)+Sa$\sqrt{\frac{1}{n_1}+\frac{1}{n_2} }* t_{n_1+n_2-2; 0.95}$

(255.8-270.74)+ 10.52*$\sqrt{\frac{1}{6} +\frac{1}{8} } * 1.782$

(-∞;-4.815)

I hope it helps!

Answer 2

The test of comparison between the mean dissolve time of each tablet can

be made using a t-test given that the sample size is small.

The correct responses are;

• a. The null hypothesis is H₀; $\overline x_1$ = $\overline x_2$, the alternative hypothesis is Hₐ; $\overline x_1$ < $\overline x_2$

• There is significant statistical evidence to suggest that the mean dissolve time is less for disk shaped tablets than for mean dissolve time for oval shaped tablets.

• b. The 95% one-sided confidence interval is; $\underline{\overline x_1 - \overline x_2 <-2.55}$

Reasons:

The given data is presented as follows;

$\begin{array}{|c|cccccc}&&Time & of & disolution\\Disk&269.0&249.3 &255.2&252.7&247.0&261.6\end{array}\right]$

$\begin{array}{|c|cccccccc}&&Time & of & disolution\\Oval&268.8&260.0&273.5&253.9&278.5&289.4&261.6&280.2\end{array}\right]$

The mean for Disks, $\overline x_1$ = 255.8

The standard deviation for Discs, s₁ ≈ 8.22

Sample size of the Disks, n₁ = 6

Mean for Oval, $\overline x_2$ ≈ 270.74

Standard deviation for Oval, s₂ ≈ 11.9

Sample size of the Oval, n₂ = 8

a. Null hypothesis is H₀; $\overline x_1$ = $\overline x_2$ (there is no difference between the mean of the samples)

Alternative hypothesis is Hₐ; $\overline x_1$ < $\overline x_2$

The standard deviation of the two populations are equal; $\sigma_{disk} = \mathbf{\sigma_{oval}}$

The pooled standard deviation, $s_p$, is given as follows;

$s_p = \mathbf{\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}}$

$s_p =\sqrt{\dfrac{\left ( 6-1 \right )\times 8.22^{2} +\left ( 8-1 \right )\times 11.9^{2}}{6+8-2}} \approx 10.524$

The test statistic is found using the following formula;

$\displaystyle t = \mathbf{ \frac{\overline x_1 - \overline x_2}{s_p \cdot \sqrt{\dfrac{1}{n_1} +\dfrac{1}{n_2} } }}$

Which gives;

$\displaystyle t = \frac{255.8 - 270.74}{10.524 \times \sqrt{\dfrac{1}{6} +\dfrac{1}{8} } } \approx -2.629$

The degrees of freedom, df = n₁ + n₂ - 2

Therefore;

df = 8 + 6 - 2  = 12

From the t-test table, we have; 0.005 < p-value < 0.01

Given that the p-value is less than the alpha level of α = 5% = 0.05, we reject the null hypothesis.

Therefore;

• There is significant statistical evidence to suggest that the mean dissolve time is less for disk shaped tablets than for oval shaped tablets.

b. The 95% one sided confidence interval is presented as follows;

$\displaystyle \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm \mathbf{t_{(\alpha /2, \, df)} \cdot s_p \cdot \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}$

$\displaystyle t_{(\alpha /2, \, df)}$ = $t_{(0.025, \, 12)}$ = 2.18

Which gives;

$\mathbf{\overline x_1 - \overline x_2} <\displaystyle \left (255.8- 270.74 \right )+2.18 \times 10.524 \cdot \sqrt{\frac{1}{6}+\frac{1}8}}$

The one sided 95% confidence interval is therefore;

• $\underline{\overline x_1 - \overline x_2 <-2.55}$

Learn more here:

https://brainly.com/question/21363975