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)`

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{(\sigma_x^2)/(n_x)+(\sigma_y^2)/(n_y)}\\\\=4.8-2.8\pm(1.96)(\sqrt{((1.9)^2)/(12)+((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)`