An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Beverage Sample Size Sample Mean Sample SD Strawberry Drink 10 530 23 Cola 10 553 16 Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P-value. (Use α = 0.05.)

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean compression strength from the extra carbonation of strawberry drink and μ2 be the mean compression strength from the extra carbonation of cola.

The random variable is μ1 - μ2 = difference in the mean compression strength from the extra carbonation of strawberry drink and the mean compression strength from the extra carbonation of cola.

We would set up the hypothesis.

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 left tailed test.

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)

From the information given,

x1 = 530

x2 = 553

s1 = 23

s2 = 16

n1 = 10

n2 = 10

t = (530 - 533) / √ (23²/10 + 16²/10)

t = - 2.6

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 = [23²/10 + 16²/10]²/[ (1/10 - 1) (23²/10) ² + (1/10 - 1) (16²/10) ²] = 6162.25/383.752

df = 16

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

p value = 0.0097

Since alpha, 0.05 > than the p value, 0.0097, then we would reject the null hypothesis. Therefore, at 5% significance level, the data suggests that the extra carbonation of cola results in a higher average compression strength.