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?

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.

Step-by-step explanation:

Data given and notation

[tex]\bar X_{A}[/tex] represent the mean for the sample A

[tex]\bar X_{B}[/tex] represent the mean for the sample B

[tex]s_{A}[/tex] represent the sample standard deviation for the sample A

[tex]s_{B}[/tex] represent the sample standard deviation for the sample B

[tex]n_{A}[/tex] sample size selected A

[tex]n_{B}[/tex] sample size selected B

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

[tex]p_v[/tex] 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:[tex]\mu_{A}-\mu_{B}= 0[/tex]

Alternative hypothesis:[tex]\mu_{A}-\mu_{B} \neq 0[/tex]

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

[tex] \bar X_A -\bar X_b \pm t_{\alpha/2} \sqrt{\frac{s^2_A}{n_A} +\frac{s^2_B}{n_B}}[/tex]

And after calculate the 90% confidence interval we got:

[tex]-4.34 < \mu_A -\mu_B < -0.03[/tex]

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.