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

Answer:

Null hypothesis:$\mu_{A}-\mu_{B}= 0$

Alternative hypothesis:$\mu_{A}-\mu_{B} \neq 0$

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

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

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.

Step-by-step 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:$\mu_{A}-\mu_{B} \neq 0$

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

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

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.