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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.
Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Respuesta :

Answer:

CI = 29.8 ± 3.53

Critical value is z = 2.58

Step-by-step explanation:

First of all let's find margin of error. It is given by the formula;

ME = zσ/√n

We are given;

Standard deviation; σ = 3.62

Sample size; n = 7

Mean; x¯ = 29.8

Now, z-value for 99% Confidence level is 2.58

Thus;

ME = (2.58 × 3.62)/√7

ME = 3.53

CI is written as;

CI = x¯ ± ME

CI = 29.8 ± 3.53

Critical value is z = 2.58

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