Answer: a) 1.029, b) (-5.318, -1.282).
Step-by-step explanation:
Since we have given that
![n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2](https://tex.z-dn.net/?f=n_1%3D13%5C%5C%5C%5Cn_2%3D17%5C%5C%5C%5C%5Cbar%7Bx_1%7D%3D1.9%5C%5C%5C%5C%5Cbar%7Bx_2%7D%3D5.2)
and
![s_1=2.1\\\\s_2=3.5](https://tex.z-dn.net/?f=s_1%3D2.1%5C%5C%5C%5Cs_2%3D3.5)
So, the standard error for comparing the means :
![SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029](https://tex.z-dn.net/?f=SE%3D%5Csqrt%7B%5Cdfrac%7Bs%5E2_1%7D%7Bn_1%7D%2B%5Cdfrac%7Bs%5E2_2%7D%7Bn_2%7D%7D%5C%5C%5C%5CSE%3D%5Csqrt%7B%5Cdfrac%7B2.1%5E2%7D%7B13%7D%2B%5Cdfrac%7B3.5%5E2%7D%7B17%7D%7D%5C%5C%5C%5CSE%3D%5Csqrt%7B1.0598%7D%5C%5C%5C%5CSE%3D1.029)
At 95% confidence interval, z = 1.96
So, Confidence interval would be
![\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)](https://tex.z-dn.net/?f=%5Cbar%7Bx_1%7D-%5Cbar%7Bx_2%7D%5Cpm%20z%5Ctimes%20SE%5C%5C%5C%5C%3D%281.9-5.2%29%5Cpm%201.96%5Ctimes%201.0294%5C%5C%5C%5C%3D-3.3%5Cpm%202.017624%5C%5C%5C%5C%3D%28-3.3-2.018%2C-3.3%2B2.018%29%5C%5C%5C%5C%3D%28-5.318%2C-1.282%29)
Hence, a) 1.029, b) (-5.318, -1.282).