Using the normal distribution and the central limit theorem, it is found that there is a 0.24 = 24% probability of observing a difference this large or larger.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
In this problem, for each sample, the mean and standard error are given by:
[tex]\mu_I = 0.4, s_I = \sqrt{\frac{0.4(0.6)}{75}} = 0.0566[/tex]
[tex]\mu_{II} = 0.4, s_{II} = \sqrt{\frac{0.4(0.6)}{90}} = 0.0516[/tex]
Hence, for the distribution of differences, the mean and the standard error are given by:
[tex]\mu = \mu_I - \mu_{II} = 0.4 - 0.4 = 0[/tex]
[tex]s = \sqrt{s_I^2 + s_{II}^2} = \sqrt{0.0566^2 + 0.0516^2} = 0.0766[/tex]
The probability of observing a difference this large or larger is given by P(|Z| > Zx), in which Zx is the z-score when X = 0.0766. Hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.09 - 0}{0.0766}[/tex]
Z = 1.175.
Hence the probability is P(|Z| > 1.175), which is 2 multiplied by the p-value of Z = -1.175. Then:
2 x 0.12 = 0.24.
0.24 = 24% probability of observing a difference this large or larger.
To learn more about the normal distribution and the central limit theorem, you can check https://brainly.com/question/24663213