Suppose that 30% of female students and 25% of male students at a large high school are enrolled in an AP ® class. Independent random samples of 20 females and 20 males are selected and asked if they are enrolled in an AP ® class. Let ˆ p F represent the sample proportion of females enrolled in an AP ® class and ˆ p F represent the sample proportion of males enrolled in an AP ® class. What is the mean of the sampling distribution of ˆ p F − ˆ p M ?

According to the central limit theorem, a sample extracted from a set of data is said to approximate a normal distribution with its sampling distribution having the same mean/proportion as the population mean/proportion.

Hence,

The sample proportion of females enrolled in an AP ® class = The population proportion of females enrolled in an AP ® class

ˆpF = 30% = 0.30

And the sample proportion of males enrolled in an AP ® class = The population proportion of males enrolled in an AP ® class

ˆpM = 25% = 0.25

So, the mean of the sampling distribution of (ˆpF − ˆpM) will simply be 30% - 25% = 5% = 0.05

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ahirohit963 ahirohit963 · Answer 2 · 2021-10-27T08:40:36+02:00

The mean of the sampling distribution of ˆpF - ˆpM is 0.05.

Given :

30% of female students and 25% of male students at a large high school are enrolled in an AP ® class.

Independent random samples of 20 females and 20 males are selected.

Solution :

Applying the central limit theorem, Sample proportion of females enrolled in an AP ® class is equal to the population proportion of females enrolled in an AP ® class

ˆpF = 30% = 0.30

And also sample proportion of males enrolled in an AP ® class is equal to the population proportion of males enrolled in an AP ® class

ˆpM = 25% = 0.25

So, the mean of the sampling distribution of ˆpF - ˆpM is,

30% - 25% = 5% = 0.05

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https://brainly.com/question/11034287