## Answers

Thursday

30/07/2020

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Therefore, a/2 = 0.025 Given that X = 106 S14 na 10 CI= 95 df = 10-1 = 9 Lower limits = X-E = 106-10.014 = 95.986 Lower Limit 95.9857 rounding above number to decimals Here, 1 Upper limit = t0.025,9 = 2.262 2 3 4 96.0 95.99 95.986 95.9857 = 106+10.014 = 116.014 Standard errors = s/n = 14/10 = 4.4272 Now, Margin of error = = ta/2.,df* std error = 2.262*4.4272 = 10.0143 C.I. = 95.986<116.014 Upper Limit 116.0143 rounding above number to decimals At 95% confidence levelt is = 1-9596 = 1 -0.95 = 0.05 Therefore, the 95% confidence interval of the population mean is = 1 2 3 116.0 116.01 116.014 116.0143 4

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