Question

An article described a study of reading speed for undergraduate and graduate students. In a sample of 12 undergraduates, the mean time to read a certain passage was 4.8 seconds, with a standard deviation of 1.9 seconds. In a sample of 24 Ph.D. students, the mean time was 2.8 seconds, with a standard deviation of 1.0 seconds. Let μXμX represent the population mean for undergraduates and let μYμY represent the population mean for Ph.D. students. Find a 95% confidence interval for the difference μX−μYμX−μY. Round down the degrees of freedom to the nearest integer and round the answers to three decimal places.

Answer 1

Answer: $(0.85,\ 3.15)$

Step-by-step explanation:

Given : $n_x=12\ ;\ \mu_x=4.8\ ;\ \sigma_x=1.9$

$n_y=24\ ;\ \mu_y=2.8\ ;\ \sigma_y=1.0$

Significance level : $\alpha: 1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Now, the confidence interval for difference of two population mean :-

$\mu_x-\mu_y\pm z_{\alpha/2}(\sqrt{\dfrac{\sigma_x^2}{n_x}+\dfrac{\sigma_y^2}{n_y}}\\\\=4.8-2.8\pm(1.96)(\sqrt{\dfrac{(1.9)^2}{12}+\dfrac{(1.0)^2}{24}}\\\\\approx2\pm1.15\\\\=(0.85,\ 3.15)$

Hence, the 95% confidence interval for the difference $\mu_x-\mu_y=(0.85,\ 3.15)$