Respuesta :
Answer:
The 95% confidence interval for the difference between the mean yields for the two types of fertilizer is approximately 92.5536 pounds < [tex]\bar{x}_{1}- \bar{x}_{2}[/tex] < 35.39 pounds
Step-by-step explanation:
The harvest from each plot treated with fertilizer A and fertilizer B are presented as follows;
Fertilizer A
445, 510, 464, 472, 441, 480, 403, 460, 448, 457, 437, 505, 417
Fertilizer B
398, 380, 368 393, 424, 387, 378, 415
The number of trials of fertilizer A, n₁ = 13
The number of trials of fertilizer B, n₂ = 8
The confidence level = 95%
α = 1 - 0.95 = 0.05
∴ α/2 = 0.025
The degrees of freedom, df = 8 - 1 = 7
[tex]t_{\alpha /2}[/tex] = 2.365
From Microsoft Excel, we get;
The standard deviation of the pounds of harvested fruits for the plots treated with fertilizer A s₁ = 30.71603
The mean of the pounds of harvested fruits for the plots treated with fertilizer A, [tex]\overline x_1[/tex] = 456.8462
The standard deviation of the pounds of harvested fruits for the plots treated with fertilizer B s = 18.99201
The mean of the pounds of harvested fruits for the plots treated with fertilizer B, [tex]\overline x_2[/tex] = 392.875
The 95% confidence interval, C.I. for the difference between the mean yields for the two types of fertilizer can be constructed with the following formula
[tex]C.I. = \left (\bar{x}_{1}- \bar{x}_{2} \right )\pm t_{\alpha /2} \cdot \sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}[/tex]
By variable substitution, we get;
[tex]C.I. = \left (456.8462- 392.875 \right )\pm 2.635 \times \sqrt{\dfrac{30.71603^{2}}{13}+\dfrac{18.99201^{2}}{8}}[/tex]
The 95% confidence interval for the difference between the mean yields for the two types of fertilizer [tex]C.I. \approx 63.9712 \pm28.5824[/tex], which can be expressed as 92.5536 lbs. < [tex]\bar{x}_{1}- \bar{x}_{2}[/tex] < 35.39 lbs.
