Apply the distributive property to factor out the greatest common factor. 44+48=
1 answer:
You might be interested in
Answer:
A) Option B is correct.
H₀: μ₁ = μ₂
Hₐ: μ₁ - μ₂ < 0
B) t = -2.502
p-value = 0.0112
C) Option A is correct.
Reject H₀. The data suggests that cola has a higher average compression strength than the strawberry drink.
D) Option A is correct.
The distributions of compression strengths are approximately normal.
Step-by-step explanation:
The complete Question is presented in the two attached images to this answer.
A) To perform this test we first define the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to test if the extra carbonation of cola results in a higher average compression strength. That is, that cola has a higher average compression strength than the strawberry drink.
Hence, the null hypothesis would be that there isn't significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.
The alternative hypothesis is that there is significant evidence to suggest that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.
Mathematically, if the average compression strength of strawberry drink is μ₁, the average compression strength of cola is μ₂ and the difference in compression strengths is μ = μ₁ - μ₂
The null hypothesis is represented as
H₀: μ = 0 or μ₁ = μ₂
The alternative hypothesis is represented as
Hₐ: μ < 0 or μ₁ - μ₂ < 0
B) So, to perform this test, we need to compute the test statistic
Test statistic for 2 sample mean data is given as
Test statistic = (μ₁ - μ₂))/σ
σ = √[(s₂²/n₂) + (s₁²/n₁)]
μ₁ = average compression strength of strawberry drink = 537
n₁ = sample size of the sample of strawberry drink in cans surveyed = 10
s₁ = standard deviation of the compression strength of strawberry drink in cans surveyed= 22
μ₂ = average compression strength of cola = 559
n₂ = sample size of the sample of cola in cans surveyed = 10
s₂ = standard deviation of the compression strength of strawberry drink in cans surveyed = 17
σ = [(17²/10) + (22²/10)] = 77.5903160379 = 8.792
We will use the t-distribution as no information on population standard deviation is provided
t = (537 - 559) ÷ 8.792
= -2.502 = -2.50
checking the tables for the p-value of this t-statistic
Degree of freedom = df = n₁ + n₂ - 2 = 10 + 10- 2 = 18
Significance level = 0.05
The hypothesis test uses a one-tailed condition because we're testing in only one direction (whether compression strength of cola in can is greater).
p-value (for t = -2.50, at 0.05 significance level, df = 18, with a one tailed condition) = 0.011154 = 0.0112 to 4 d.p.
C) The interpretation of p-values is that
When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.
So, for this question, significance level = 0.05
p-value = 0.0112
0.0112 < 0.05
Hence,
p-value < significance level
This means that we reject the null hypothesis accept the alternative hypothesis & say that the extra carbonation of cola results in a higher average compression strength, that is, cola has a higher average compression strength than the strawberry drink.
D) The necessary conditions required before a t-test is deemed valid include.
- The samples used must be a random sample of the population distribution with each variable in the sample independent of other one.
- The distribution of the population where the samples were extracted from must be normal or approximately normal to ensure some degree of normality for the samples.
Hence, the necessary assumption for this t-test among the options is that the distributions of compression strengths are approximately normal.
Hope this Helps!!!
