a sample of 25 accounts receivable and determined the mean of the sample to of $

Question

a sample of 25 accounts receivable and determined the mean of the sample to of $500. She reported that the sample information indicated the . A statistici ian selected be $3,000 wit mean of h a standard deviation the population rang es from $2793.61 to $3206.39. She did not report what confidence le nad used. Based on the above information, determine the confidence coefficient that was used. [15 points] a. Find the mean of the confidence interval. b. Find the margin of error. c. Using this information find the confidence coefficient.

Explanation / Answer

a) mean = 1/2*(2793.61 + 3206.39) = 3000
b) margin or error = 206.39
c) t * s/sqrt(n) = 206.39
here n = 25
, s = 500
hence
t = 206.39/(500/sqrt(25))
= 2.0639

df= n-1 = 24

P(T > 2.0369) = 0.025

hence 95% confidence interval

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