Suppose a random sample of n measurements is selected from a binomial population with probability of success p = .38. Given n = 300, describe the shape, and find the mean and the standard deviation of the sampling distribution of the sample proportion, .
The Sampling Distribution of the Sample Proportion. For large samples, the sample proportion is approximately normally distributed, with mean μˆP=p. and standard deviation σˆP=√pqn.
1) for large samples (n=300) the sample proportion is approximately normally distributed,
Hence,
The Shape of sampling distribution of the sample proportion is symmetrical.( The shape of normal distribution is symmetrical)
Given : p = 0.38 , n = 300
2) Mean = p = probability of success
Mean of the sampling distribution of the sample proportion is p = 0.38
Hence mean is 0.38
3) Standard deviation = √pqn
p= 0.38
q=1-p =1-0.38 = 0.62
n =300
Variance is npq = 0.38*0.62*300 = 70.68
Standard deviation of the sampling distribution of the sample proportion is √pqn =√0.38*0.62*300 =8.407139822793480
Hence, Standard deviation is ✓variance = 8.407139822793480
Standard deviation is 8.4071 (to 4 decimal places)
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