## Answers

Solution: The provided sample mean is X = 39.04 and the sample standard deviation is s = 11.51. The size of the sample is n = 83 and the required confidence level is 95%. The number of degrees of freedom are df = 83 – 1 = 82, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 82 degrees of freedom is te = 1.989 The 95% confidence for the population mean u is computed using the following expression C1=(x - to + ) Therefore, based on the information provided, the 95% confidence for the population mean u is 1.989 x 11.51 ,39.04 + 1.989 x 11.51 CI= ( 39.04 - 83 783 ) = (39.04 – 2.513, 39.04 +2.513) = (36.527, 41.553)

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