Mathematics, 27.02.2020 21:35 RicoGonza

Given two dependent random samples with the following results: Population 1 71 68 50 84 76 76 80 79 Population 2 76 63 54 80 79 82 75 82 Can it be concluded, from this data, that there is a significant difference between the two population means? Let d=(Population 1 entry)−(Population 2 entry). Use a significance level of α=0.2 for the test. Assume that both populations are normally distributed. State the null and alternative hypotheses for the test.

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Answer from: Gimagg8195

$t=\frac{73-73.875}{\sqrt{\frac{(10.570)^2}{8}+\frac{(10.106)^2}{8}}}}=-0.169$

$df=n_{1}+n_{2}-2=8+8-2=14$

Since is a bilateral test the p value would be:

$p_v =2*P(t_{(14)}$

So the p value is a very high value and using any significance level given $\alpha=0.2$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the means are similar for both groups

Step-by-step explanation:

Data given and notation

1:[71 68 50 84 76 76 80 79]

2:[76 63 54 80 79 82 75 82]

$\bar X_{1}=73$ represent the mean for the sample 1

$\bar X_{2}=73.875$ represent the mean for the sample 2

$s_{1}=10.570$ represent the sample standard deviation for the sample 1

$s_{2}=10.106$ represent the sample standard deviation for the sample 2

$n_{1}=8$ sample size for the group 1

$n_{2}=8$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for both groups are equal , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are less than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{73-73.875}{\sqrt{\frac{(10.570)^2}{8}+\frac{(10.106)^2}{8}}}}=-0.169$

Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case: