Question

Scientists want to estimate the mean weight of mice after they have been fed a special diet. From previous studies, it is known that the weight is normally distributed with standard deviation 5 grams. How many mice must be weighed so that a 95% confidence interval will have margin of error of 0.6 grams? Write only an integer as your answer.

Answer 1

Solution:- Given that S = 5, E = 0.6, 95% Confidence interval = 1.96

n = (Z*s/E)^2 = (1.96*5/0.6)^2 = 266.7777

n = 267

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Solution:- Given that samples : 19.1,20.1,20.8,22.0,20.5,19.5
X = 20.333 , s = 1.029, n = 6, df = n-1 = 6-1 = 5
95% Confidence interval

95% Confidence interval for the true mineral : (19.253,21.413)

The provided sample mean is X = 20.333 and the sample standard deviation is s = 1.029. The size of the sample is n = 6 and the required confidence level is 95%. The number of degrees of freedom are df = 6-1 = 5, and the significance level is a = 0.05. Based on the provided information, the critical t-value for a = 0.05 and df = 5 degrees of freedom is te = 2.571. The 95% confidence for the population mean u is computed using the following expression CI = x = (* te XS 2 te x 8 - 72 Therefore, based on the information provided, the 95% confidence for the population mean is CI = 20.333 2.571 x 1.029 V6 20.333 + 2.571 x 1.029 V6 (20.333 - 1.08, 20.333 + 1.08) = (19.253, 21.413) which completes the calculation