Using the t-distribution, it is found that the 98% confidence interval estimate for the difference between the population means of the first sample and of the second sample is (2.8, 3.4). Since the interval is entirely positive, it means that treatment appears to reduce the mean amount of metal removed.
The standard errors for each sample are given by:
[tex]s_1 = \frac{1.1}{\sqrt{100}} = 0.11[/tex]
[tex]s_2 = \frac{0.9}{\sqrt{200}} = 0.0636[/tex]
The distribution of the difference has:
[tex]\overline{x} = \mu_1 - \mu_2 = 12.2 - 9.1 = 3.1[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.11^2 + 0.0636^2} = 0.1271[/tex]
The confidence interval is:
[tex]\overline{x} \pm ts[/tex]
The critical value, using a t-distribution calculator, for a two-tailed 98% confidence interval, with 100 + 200 - 2 = 298 df, is t = 2.3389.
Hence, the interval is:
[tex]\overline{x} - ts = 3.1 - 2.3389(0.1271) = 2.8[/tex]
[tex]\overline{x} + ts = 3.1 + 2.3389(0.1271) = 3.4[/tex]
The 98% confidence interval estimate for the difference between the population means of the first sample and of the second sample is (2.8, 3.4). Since the interval is entirely positive, it means that treatment appears to reduce the mean amount of metal removed.
A similar problem is given at https://brainly.com/question/15180581
