Question

A meat inspector has randomly selected 25 packs of 98% lean beef produced by Company A. Thesample resulted in a mean of 96% with a sample standard deviation of 1%. Assume normality.

(1) Find a 99% confidence interval for the mean leanness of lean beef produced by company A.

(2) Find a 95% prediction interval for the leanness of a new pack.

Answer 1

(1)

The provided sample mean is X = 0.96 and the sample standard deviation is s = 0.01. The size of the sample is n=25 and the required confidence level is 99%. The number of degrees of freedom are df = 25 – 1 = 24, and the significance level is a = 0.01. Based on the provided information, the critical t-value for a = 0.01 and df = 24 degrees of freedom is te = 2.797. The 99% confidence for the population mean ji is computed using the following expression CI = x te x 8 vn X + te x 8 in Therefore, based on the information provided, the 99 % confidence for the population mean pe is 2.797 x 0.01 CI = 0.96 (0.96 0.96 + 2.797 x 0.01 25 V25 = (0.96 -0.006, 0.96 +0.006) = = (0.954, 0.966)

  $\\ We\ are\ 99\%\ confident\ that\ \\ \\ 95.4\%<\mu<69.6\%$

(b)

$\\ 95\%\ prediction\ interval\ is:\\ \\ \left(\bar{x}-t_{\alpha}s\left(\sqrt{1+\frac{1}{n}} \right ),\bar{x}+t_{\alpha}s\left(\sqrt{1+\frac{1}{n}} \right )\right)\\ \\ \\ \left(96- 2.064*1\left(\sqrt{1+\frac{1}{25}} \right ),96+2.064*1\left(\sqrt{1+\frac{1}{25}} \right )\right)\\ \\ \\ (96-2.1049,96+2.1049)\\ \\ \\ (93.8951,98.1049)$