A random sample of 4"4 fields" of barley has a mean yield of "49.6"49.6 bushels per acre and standard deviation of 7.997.99 bushels per acre. Determine the 99%99% confidence interval for the true mean yield. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

The critical value that should be used in constructing the interval is T = 5.8408.

The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the students t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 4 - 1 = 3

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of . So we have T = 5.8408. This is the critical value.

The margin of error is:

M = T*s = 5.8408*7.99 = 46.668 bushels per acre

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 49.6 - 46.668 = 2.943 bushels per acre.

The upper end of the interval is the sample mean added to M. So it is 49.6 + 46.668 = 96.268 bushels per acre.

The 99% confidence interval for the true mean yield is between 2.943 bushels per acre and 96.268 bushels per acre.

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation:

step-by-step explanation:   divide the numerator by the denominator to get the decimal

it would still be 193

if divide by 100 then 1.93

i am not sure