Answer:
Step-by-step explanation:
Given that sample size is 130 >30. Also by central limit theorem, we know that mean (here proportion) of all means of different samples would tend to become normal with mean = average of all means(here proportions)
Hence we can assume normality assumptions here.
Proportion sample given = 92/130 = 0.7077
The mean proportion of different samples for large sample size will follow normal with mean = sample proportion and std error = square root of p(1-p)/n
Hence mean proportion p= 0.7077
q = 1-p =0.2923
Std error = 0.0399
For 95% confidence interval we find that z critical for 95% two tailed is 1,.96
Hence margin of error = + or - 1.96(std error)
= 0.0782
Confidence interval = (p-margin of error, p+margin of error)
= (0.7077-0.0782,0.7077+0.0782)
=(0.6295, 0.7859)
We are 95% confident that average of sample proportions of different samples would lie within these values in the interval for large sample sizes.
