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According to a recent survey, women chat on their mobile phones more than do men (CNN, August 25, 2010). Irina wants to estimate the difference between the average chat time for female and male college students. She takes a random sample of 40 male students and 40 female students in her college. She finds that female students chatted for a sample average of 820 minutes per month with a sample standard deviation of 160 minutes. Male students, on the other hand, chatted for a sample average of 760 minutes per month with a sample standard deviation of 240 minutes. You may assume that the population variances are equal. (a) Construct a 95% confidence interval for the difference between the population means.

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Answer:

[tex](820-760) -1.99 \sqrt{\frac{160^2}{40}+ \frac{240^2}{40}} =-30.758[/tex]

[tex](820-760) +1.99 \sqrt{\frac{160^2}{40}+ \frac{240^2}{40}} =150.758[/tex]

Step-by-step explanation:

For this case we have the following data given:

[tex]\bar X_{m}= 760[/tex] the sample mean for males

[tex]\bar X_{f}= 820[/tex] the sample mean for females

[tex]s_m = 240[/tex] deviation for males

[tex]s_f = 160[/tex] deviation for females

[tex]n_m = 40[/tex] sample size of males

[tex]n_f = 40[/tex] sample size of females

Confidence interval

The confidence interval for the true mean differences is given by:

[tex] (\bar_{f} -\bar_m) \pm t_{\alpha/2} \sqrt{\frac{s^2_f}{n_f} +\frac{s^2_m}{n_m}}[/tex]

The degrees of freedom are given by:

[tex] df = n_f +n_m -2= 40+40-2= 78[/tex]

The confidence level is 0.95 or 95% the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 = 0.025[/tex] and the critical value for this case is:

[tex]t_{\alpha/2} =1.99[/tex]

And replacing we got:

[tex](820-760) -1.99 \sqrt{\frac{160^2}{40}+ \frac{240^2}{40}} =-30.758[/tex]

[tex](820-760) +1.99 \sqrt{\frac{160^2}{40}+ \frac{240^2}{40}} =150.758[/tex]

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