Thirty middle-aged women with serum triiodothyronine (T3) readings between 180 nanograms and 200 nanograms per deciliter selected randomly from a population of similar female patients at a large local hospital. Fifteen of the 30 women were assigne d group A and received a placebo. The other 15 women were assigned to group B and received a new T3 drug. After three months, posttreatment T3 readings were taken for all 30 women and were compared with pretreatment readings. The reduction in T3 level (Pretreatment reading - Posttreatment reading) for each woman in the study is shown here.

Group A (placebo

reduction (in nanograms per deciliter): 2, 19, 8, 4, 12, 8, 17, 7, 24, 1, 3, 7, 11, 15, 9

Group B (T3 drug reduction (in nanograms per deciliter): 30, 19, 18, 17, 20, -4, 23, 10, 9, 22, 17, 21, 13, 31, 15

Part A: Do the data provide convincing evidence, at the a = 0.01 level, that the T3 drug is effective in producing a reduction in mean T3 level? (6

points) Part B: Create and interpret a 99% confidence interval for the difference in the placebo and the new drug (4 points)

[tex]\begin{array}{ccc}Group \, A \ (Placebo)& Group \, B \ T3 \ drug \ reduction \ (ng/dL)\\2&30\\19&19\\8&18\\4&17\\12&20\\ 8&-4\\17&23\\7&10\\24&9\\1&22\\3&17\\7&21\\11&13\\15&31\\9&15\\\end{array}[/tex]

The mean for Group A, [tex]\overline x_1[/tex] = 9.8

The standard deviation for Group A, s₁ ≈ 6.614

The number of members of Group A, n₁ = 15

The mean for Group B, [tex]\overline x_2[/tex] = 17.4

The standard deviation for Group B, s₂ ≈ 8.56738

The number of members of Group A, n₂ = 15

The confidence level, α = 0.01 level

The null hypothesis, H₀: [tex]\overline x_1[/tex] ≤ [tex]\overline x_2[/tex]

The alternative hypothesis, Hₐ: [tex]\overline x_1[/tex] > [tex]\overline x_2[/tex]

The t-test formula as follows;

[tex]t=\dfrac{\overline x_{1}-\overline x_{2}}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}[/tex]

Plugging in the values of the variables, we get;

[tex]t=\dfrac{ 17.4 - 9.8}{\sqrt{\dfrac{8.56738^{2}}{15} - \dfrac{6.614^{2} }{15}}} \approx 5.405[/tex]

The critical 't' value at 0.01 convincing level and df = 15 - 1 = 14 is 2.624

Therefore, given that the t value is larger than the critical 't', we reject the null hypothesis and we fail to reject the alternative hypothesis, and there is convincing evidence that at a confidence level of a = 0.01, the T3 drug is effective in producing a reduction in mean T3 level

Part B

The 99% confidence interval is given by the following formula;

[tex]\left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}[/tex]

[tex]t_{\alpha /2}[/tex] = 2.977

Plugging in the values, we get;

[tex]C.I. = \left (9.8- 17.4\right )\pm 2.977 \times \sqrt{\dfrac{6.614^{2}}{15}+\dfrac{8.56738 ^{2}}{15}}[/tex]

The 99% confidence interval for the difference in the placebo and the new drug is C.I. = -15.92 < [tex]\overline x_1[/tex] - [tex]\overline x_2[/tex] < 0.7195