Respuesta :
Using the z-distribution, it is found that the 98% confidence interval for the difference in the proportion of apples with blemishes is of (-0.1055, 0.2655).
Sample proportions and standard errors
The sample proportions, and standard error for each sample, are given as follows:
[tex]p_G = \frac{48}{75} = 0.64[/tex]
[tex]s_G = \sqrt{\frac{0.64(0.36)}{75}} = 0.0554[/tex]
[tex]p_R = \frac{42}{75} = 0.56[/tex]
[tex]s_R = \sqrt{\frac{0.56(0.44)}{75}} = 0.0573[/tex]
Distribution of differences
The distribution of differences has mean and standard error given by:
[tex]p = p_G - p_R = 0.64 - 0.56 = 0.08[/tex]
[tex]s = \sqrt{s_G^2 + s_R^2} = \sqrt{0.0554^2 + 0.0573^2} = 0.0797[/tex]
What is the confidence interval?
The confidence interval is given by:
[tex]p \pm zs[/tex]
z is the critical value. We have a 98% confidence level, hence, using a z-distribution calculator, it is of z = 2.327.
The lower limit of this interval is:
[tex]\pi - zs = 0.08 - 2.327(0.0797) = -0.1055[/tex]
The upper limit of this interval is:
[tex]\pi + zs = 0.08 + 2.327(0.0797) = 0.2655[/tex]
The 98% confidence interval for the difference in the proportion of apples with blemishes is of (-0.1055, 0.2655).
More can be learned about the z-distribution at https://brainly.com/question/25890103
