Question

The null and alternate hypotheses are: H0 : μ1 = μ2 H1 : μ1 ≠ μ2 A random sample of 15 observations from the first population revealed a sample mean of 350 and a sample standard deviation of 12. A random sample of 17 observations from the second population revealed a sample mean of 342 and a sample standard deviation of 15. At the 0.10 significance level, is there a difference in the population means?

Answer 1

Explanation:

From the information given,

The formula for determining the t test statistic is

t = (x1 - x2)/√(s1²/n1 + s2²/n2)

Where

x1 = sample mean of the first population

x2 = sample mean of the second population

s1 = sample standard deviation of the first population

s2 = sample standard deviation of the second population

n1 = number of samples in the first population

n2 = number of samples in the second population

From the information given,

x1 = 350

x2 = 342

s1 = 12

s2 = 15

n1 = 15

n2 = 17

Therefore,

t = (350 - 342)/√(12²/15 + 15²/17)

= 8/√(9.6 + 13.24)

= 1.674

degree of freedom is

n2 + n1 - 2 = 17 + 15 - 2 = 30

The test is 2 tailed. Using the test statistic calculator, the probability value is 0.105

Comparing the probability value with the significance level, there is no significant difference. Therefore, we would fail to reject the null hypothesis

Answer 2

`t=\frac{350-342}{\sqrt{(12^2)/(15)+(15^2)/(17)}}}=1.674`

`p_v =2*P(t_((20))>1.674)=0.105`

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there are not ignificant differences at 10% of significance.

Explanation:

Data given and notation

`\bar X_(1)=350` represent the mean for the sample 1

`\bar X_(2)=342` represent the mean for the sample 2

`s_(1)=12` represent the sample standard deviation for 1

`s_(2)=15` represent the sample standard deviation for 2

`n_(1)=15` sample size selected for 1

`n_(2)=10` sample size selected for 2

`\alpha=0.1` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the means are different or no, the system of hypothesis would be:

Null hypothesis:
`\mu_(1) = \mu_(2)`

Alternative hypothesis:
`\mu_(1) \\eq \mu_(2)`

If we analyze the size for the samples both are less than 30 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{(s^2_(1))/(n_(1))+(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".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=\frac{350-342}{\sqrt{(12^2)/(15)+(15^2)/(17)}}}=1.674`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n_(1)+n_(2)-2=15+17-2=30`

Since is a two sided test the p value would be:

`p_v =2*P(t_((20))>1.674)=0.105`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and there are not ignificant differences at 10% of significance.