The researcher obtains the scores on the test from a random sample of 16 users of StudyFocus and a random sample of 15 users of Prepara. (These samples are chosen independently.) For the users of StudyFocus, their sample mean is 490.8 with a sample variance of 5932.5. For the users of Prepara, their sample mean is 499.7 with a sample variance of 755.8.
Assume that the two populations of scores are approximately normally distributed. Let μ1 be the population mean score on the test by the users of StudyFocus. Let μ2 be the population mean score on the test by the users of Prepara.
Construct a 99% confidence interval for the difference μ1−μ2. Then find the lower and upper limit of the 99% confidence interval. Carry your intermediate computations to three or more decimal places. Round your answers to two or more decimal places. (If necessary, consult a list of formulas.)
Lower limit: ______
Upper limit: ______
Answer:
Given Information in question:
sample mean 1 =490.8
sample mean 2 =499.7
sample standard deviation 1 =77.02
sample standard deviation 2 =27.49
sample size 1 =16
sample size 2 =15
confidence interval level = 99%
Solution:
☻ The level of significance $(\alpha) = 1 – 0.99 = 0.01 $
☻ The degree of freedom $(df):$
$alpha = 1 – \text{the confidence interval level} = 1 – 0.99 = 0.01$
☻ The critical value $(t_c)$
$t_c = t_{\frac{\alpha}{2}, \text{df}} = t_{0.01/2, 19.512} = 2.853$
☻ The confidence interval:
$CI = (\overline{x}_1 – \overline{x}_2) \pm t_c \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = (490.8 – 499.7) \pm 2.853 \sqrt{\frac{77.02^2}{16} + \frac{27.49^2}{15}} = (-67.443, 49.643)$
