Question

A researcher was interested in comparing the heights of women in two different countries. Independent simple random samples of 9 women from country A and 9 women from country B collected data for their heights (in inches). The following 90% confidence interval was obtained for the difference between the mean height of women in country A and the mean height of women in country B. -4.34 in. < μA - μB < -0.03 in What does the confidence interval suggest about the population means?

Answer 1

Null hypothesis:
`\mu_(A)-\mu_(B)= 0`

Alternative hypothesis:
`\mu_(A)-\mu_(B) \\eq 0`

And for this case we can use the confidence interval given by:

`\bar X_A -\bar X_b \pm t_(\alpha/2) \sqrt{(s^2_A)/(n_A) +(s^2_B)/(n_B)}`

And after calculate the 90% confidence interval we got:

`-4.34 < \mu_A -\mu_B < -0.03`

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.

Explanation:

Data given and notation

`\bar X_(A)` represent the mean for the sample A

`\bar X_(B)` represent the mean for the sample B

`s_(A)` represent the sample standard deviation for the sample A

`s_(B)` represent the sample standard deviation for the sample B

`n_(A)` sample size selected A

`n_(B)` sample size selected B

`\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 two means are equal., the system of hypothesis would be:

Null hypothesis:
`\mu_(A)-\mu_(B)= 0`

Alternative hypothesis:
`\mu_(A)-\mu_(B) \\eq 0`

And for this case we can use the confidence interval given by:

`\bar X_A -\bar X_b \pm t_(\alpha/2) \sqrt{(s^2_A)/(n_A) +(s^2_B)/(n_B)}`

And after calculate the 90% confidence interval we got:

`-4.34 < \mu_A -\mu_B < -0.03`

So then we see that all the values are negative so then we can conclude that the two means are different and on this case the mean for A seems to be lower than the mean for B at 10% of significance.