The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. Step 1 of 2: Suppose a sample of 2845 tenth graders is drawn. Of the students sampled, 2276 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. Enter your answer as a fraction or a decimal number rounded to three decimal places.
Step 2 of 2: Suppose a sample of 2845 tenth graders is drawn. Of the students sampled, 2276 read above the eighth grade level. Using the data, construct the 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.

Step 1: The estimate the proportion of tenth graders reading at or below the eighth grade level is 0.2.

Step 2: The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

Suppose a sample of 2845 tenth graders is drawn. Of the students sampled, 2276 read above the eighth grade level.

So 2845 - 2276 = 569 read below, and the estimate of the proportion of tenth graders reading at or below the eighth grade level is:

, and the answer to step 1 is 0.2.

The sample size is

99% confidence level

So , z is the value of Z that has a p-value of , so .  

The lower limit of this interval is:

The upper limit of this interval is:

The 99% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.181, 0.219).

not sure but it may be b, as i said i'm not sure.

8/k

step-by-step explanation: