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.

details

step-by-step explanation:

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation: