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? Conduct a hypothesis test at the
α = 5% level.
a. State the appropriate null and alternative hypotheses.
b. Compute the test statistic.
c. Compute the P-value.
d. State the conclusion of the test in the context of this setting.

Step-by-step explanation:

This is a test of 2 independent groups. Let μ1 be the mean dissolution time for disk-shaped ibuprofen tablets and μ2 be the mean dissolution time for oval-shaped ibuprofen tablets.

The random variable is μ1 - μ2 = difference in the mean dissolution time for disk-shaped ibuprofen tablets and the mean dissolution time for oval-shaped ibuprofen tablets.

We would set up the hypothesis.

a) The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

This is a two tailed test.

For disk shaped,

Mean, x1 = (269.0 + 249.3 + 255.2 + 252.7 + 247.0 + 261.6)/6 = 255.8

Standard deviation = √(summation(x - mean)²/n

n1 = 6

Summation(x - mean)² = (269 - 255.8)^2 + (249.3 - 255.8)^2 + (255.2 - 255.8)^2+ (252.7 - 255.8)^2 + (247 - 255.8)^2 + (261.6 - 255.8)^2 = 337.54

Standard deviation, s1 = √(337.54/6) = 7.5

For oval shaped,

Mean, x2 = (268.8 + 260 + 273.5 + 253.9 + 278.5 + 289.4 + 261.6 + 280.2)/8 = 270.7375

n2 = 8

Summation(x - mean)² = (268.8 - 270.7375)^2 + (260 - 270.7375)^2 + (273.5 - 270.7375)^2+ (253.9 - 270.7375)^2 + (278.5 - 270.7375)^2 + (289.4 - 270.7375)^2 + (261.6 - 270.7375)^2 + (280.2 - 270.7375)^2 = 991.75875

Standard deviation, s2 = √(991.75875/8) = 11.1

b) Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

Therefore,

t = (255.8 - 270.7375)/√(7.5²/6 + 11.1²/8)

t = - 3

c) The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [7.5²/6 + 11.1²/8]²/[(1/6 - 1)(7.5²/6)² + (1/8 - 1)(11.1²/8)²] = 613.86/51.46

df = 12

We would determine the probability value from the t test calculator. It becomes

p value = 0.011

d) Since alpha, 0.05 > than the p value, 0.011, then we would reject the null hypothesis. Therefore, we can conclude that at 5% significance level, the mean dissolve times differ between the two shapes

for b its (t) and c its (d)

step-by-step explanation:

nope.   happens to me as well.

-aparri