2. in a random sample of 130 world campus students, 92 were employed full-time. we want use these data to construct a 95% confidence interval to estimate the proportion of all world campus students who are employed full-time. is it appropriate to use the normal approximation method? show how you checked assumptions.

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.

add all the numbers and check the ones place.

the answer is 486. as the ones place is more than 4, the estimation will be the next 10 ie, 490

i don't think there is enough