The United States government wanted to determine what proportion of Americans approve of the current President so they conducted a survey of 2,000 randomly chosen Americans. Assume that the true proportion of Americans who approve of the president is 0.46. Determine what the mean and the standard deviation of the sampling distribution of the sample proportion would be. Round to 2 decimal places when necessary. The mean of the sampling distribution is .43 and the standard deviation of the sampling distribution is .05 .

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

[tex]p = 0.46, n = 2000[/tex]

So

Mean: [tex]\mu = p = 0.46[/tex]

Standard deviation: [tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.46*0.54}{2000}} = 0.01[/tex]

The mean of the sampling distribution is 0.46

The standard deviation of the sampling distribution is 0.01

abidemiokin abidemiokin · Answer 2 · 2022-01-14T12:43:21+01:00

The mean and the standard deviation of the sampling distribution of the sample proportion are 0.46 and 0.01 respectively

Proportion and standard deviation

The formula for calculating the standard deviation for a sample proportion is given as:

[tex]\sigma =\sqrt{\frac{p(1-p)}{n} }[/tex]

p is the proportion
n is the sample size

Note that p = mean = 0.46

To get the standard deviation:

[tex]\sigma =\sqrt{\frac{0.46(1-0.46)}{2000} }\\\sigma =\sqrt{\frac{0.46(0.54)}{2000} }\\\sigma = 0.01[/tex]

Hence the mean and the standard deviation of the sampling distribution of the sample proportion are 0.46 and 0.01 respectively.

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