Question

PLEASE DOUBLE CHECK ANSWER, AND NO HANDWRITTEN ANSWERS. THANK YOU!!

A random sample of 7 fields of durum wheat has a mean yield of 29.8 bushels per acre and standard deviation of 3.62 bushels per acre. Determine the 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.

Step 2 of 2: Construct the 99% confidence interval. Round your answer to one decimal place.

Answer #1

Solution :

Given that,

Point estimate = sample mean = = 29.8

sample standard deviation = s = 3.62

sample size = n = 7

Degrees of freedom = df = n - 1 = 6

t_{
/2,df} = t_{0.005,6} = 3.707

Margin of error = E = t_{/2,df}
* (s /n)

= 3.707 * (3.62 / 7)

Margin of error = E = 5.1

The 99% confidence interval estimate of the population mean is,

- E < < + E

29.8 - 5.1 < < 29.8 + 5.1

24.7 < < 34.9

(24.7 , 34.9)

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