Which is not in the form of a ratio
1 answer:
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Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity
Sample 1 (aqueous film forming foam)
n₁= 5
X[bar]₁= 4.7
S₁= 0.6
Sample 2 (alcohol-type concentrates )
n₂= 5
X[bar]₂= 6.8
S₂= 0.8
Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?
The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:
t=
a) 95% CI
(X[bar]_1 - X[bar]_2) ± *
Sa²= =
= 0.5
Sa= 0.707ç
(4.7-6.9) ± 2.306*
[-4.78; 0.38]
With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.
b.
The hypothesis is:
H₀: μ₁ - μ₂= 0
H₁: μ₁ - μ₂≠ 0
α: 0.05
The interval contains the cero, so the decision is to reject the null hypothesis.
<u>Complete question</u>
a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.
b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z= where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z= =-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.