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A researcher caught 25 wild Rattatas on UI campus and found theiraverage weight is 7.8lbs. Find the 95% confidence interval of the population mean weight of Rattatas. The population variance is unknown and the researcher estimated the sample vairance to be S^2 =5.

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The provided sample mean is X 7.8 and the sample standard deviation is s 2.5. The size of the sample is n = 25 and the required confidence level is 95% The number of degrees of freedom are df = 25-1 = 24, and the significance level is α = 0.05. Based on the provided information, the critical t-value for a 0.05 and df 24 degrees of freedonm is tc 2.064. The 95% confidence for the population mean μ is computed using the following expression Therefore, based on the information provided, the 95 % confidence for the population mean μ is 2.064-25, 7,8 2.064 x 2.5 丽 CI- 7.8 - (7.8 1.032,7.8+1.032) = (6.768, 8.832) which completes the calculation.

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