Question

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	15	532	21
Cola	15	554	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.)

State the relevant hypotheses. (Use μ₁ for the strawberry drink and μ₂ for the cola.)

H₀: μ₁ − μ₂ = 0
H_a: μ₁ − μ₂ > 0H₀: μ₁ − μ₂ = 0
H_a: μ₁ − μ₂ ≠ 0    H₀: μ₁ − μ₂ = 0
H_a: μ₁ − μ₂ ≥ 0H₀: μ₁ − μ₂ = 0
H_a: μ₁ − μ₂ < 0

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)

t	=
P-value	=

State the conclusion in the problem context.

Reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.

Reject H₀. The data does not suggest that cola has a higher average compression strength than the strawberry drink.   

Fail to reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.

Fail to reject H₀. The data does not suggest that cola has a higher average compression strength than the strawberry drink.

What assumptions are necessary for your analysis?

The distributions of compression strengths are the same.

The distributions of compression strengths have equal variances.    

The distributions of compression strengths are approximately normal.

The distributions of compression strengths have equal means.

Answer 1

n1 = 15

= 532

s1 = 21

s1^2 = 441

n2 = 15

= 554

s2 = 16

s2^2 = 256

Claim: The extra carbonation of cola results in a higher average compression strength.

The null and alternative hypothesis is

H0: μ1 − μ2 = 0
Ha: μ1 − μ2 > 0

For doing this test first we have to check the two groups have population variances are equal or not.

The null and alternative hypothesis is

Test statistic is

F = largest sample variance / Smallest sample variances

F = 441 / 256 = 1.72

Degrees of freedom => n1 - 1 , n2 - 1 => 15 - 1 , 15 - 1 => 14 , 14

Critical value = 2.484 ( Using f table)

Critical value > test statistic so we fail to reject null hypothesis.

Conclusion: The population variances are equal.

So we have to use here pooled variance.

Test statistic is

Degrees of freedom = n1 + n2 - 2 = 15 + 15 - 2 = 28

P-value = P(T < - 3.23) = 0.002

P-value < 0.05 we rejthe ect null hypothesis.

Conclusion: Reject H0. The data suggest that cola has a higher average compression strength than the strawberry drink.

Assumptions:

The distributions of compression strengths have equal variances.    

The distributions of compression strengths are approximately normal.