STUN. Herwinyes 8C3/8C4 Elementary Statistics laps Elvira Jinadu & I 07/30/20 11:31 PM Homework: Section 6.2...

solution:

6.2.25-T

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6.2.26-T

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The sample mean X is computed as follows: i ==> X = 10931 12 910.917 i=1

Also, the sample variance s is o==+(*-:(x)) - (197876 Mogna = 92849.538 Therefore, the sample sandartd deviation s is s= V s2 = 92849.538 = 304.712

9,8+ the mome) The provided sample mean is Ħ = 910.917 and the sample standard deviation is s = 304.712. The size of the sample is n = 12 and the required confidence level is 90%. The number of degrees of freedom are df = 12 - 1=11, and the significance level is a = 0.1. Based on the provided information, the critical t-value for a = 0.1 and df = 11 degrees of freedom is te = 1.796. The 90% confidence for the population mean is computed using the following expression CI = x te xs Vn Therefore, based on the information provided, the 90 % confidence for the population mean u is CI = ( 910.917 1.796 x 304.712 -,910.917+ 12 1.796 x 304.712 12 (910.917 - 157.971,910.917 + 157.971) = (752.946, 1068.888)

The sample mean X is computed as follows: 27.9 12 2.325 1=1

Also, the sample variance sº is $? +($*-:(**)) - 2 (1995-1990) -- = 1.408 n Therefore, the sample sandartd deviation s is s=32 = V1.408 = 1.186

The provided sample mean is X = 2.325 and the sample standard deviation is s = 1.186. The size of the sample is n = 12 and the required confidence level is 99%. The number of degrees of freedom are df = 12 - 1=11, and the significance level is a = 0.01. Based on the provided information, the critical t-value for a = 0.01 and df = 11 degrees of freedom is te = 3.106 The 99% confidence for the population mean pe is computed using the following expression te x 8 CI = x te XS (* X + vn Therefore, based on the information provided the 99 % confidence for the population mean u is 3.106 x 1.186 3.106 x 1.186 CI = (2.325 ,2.325+ V12 V12 = (2.325 - 1.063, 2.325 +1.063) = (1.262, 3.388)