Respuesta :
Answer:
[tex](31.1 -30.4) - 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.659[/tex]
[tex](31.1 -30.4) + 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.741[/tex]
So then the 99% confidence interval for [tex] \mu_X -\mu_Y[/tex] is givenby (0.659; 0.741)
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
We have the following data given:
[tex] \bar X = 31.1[/tex] represent the sample mean for women
[tex]n_x = 1920[/tex] the sample size of women
[tex]s_x = 0.2[/tex] standard deviation for women
[tex] \bar Y = 30.4[/tex] represent the sample mean for men
[tex]n_y = 1551[/tex] the sample size of men
[tex]s_y = 0.6[/tex] standard deviation for men
Solution to the problem
The confidence interval is given by the following formula:
[tex](\bar X -\bar Y) + t_{\alpha/2} \sqrt{\frac{s^2_x}{n_x}+\frac{s^2_y}{n_y}}[/tex] (1)
Since the sample sizes are large enough we can use the normal distribution as approximation of the t distribution. For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex] z_{\alpha/2}=2.58[/tex]
And replacing into the formula (1) we got:
[tex](31.1 -30.4) - 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.659[/tex]
[tex](31.1 -30.4) + 2.58 \sqrt{\frac{0.2^2}{1920}+\frac{0.6^2}{1551}}= 0.741[/tex]
So then the 99% confidence interval for [tex] \mu_X -\mu_Y[/tex] is givenby (0.659; 0.741)
