Question #340287

the average grade of 30 students in their statistics and probability examination is 85 with a standard deviation of 5. find the interval for the true mean at 95% interval


1
Expert's answer
2022-05-13T05:14:51-0400

The critical value for α=0.05\alpha = 0.05 and df=n1=29df = n-1 = 29 degrees of freedom is tc=z1α/2;n1=2.04523.t_c = z_{1-\alpha/2; n-1} =2.04523.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉ+tc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})

=(852.04523×530,85+2.04523×530)=(85-2.04523\times\dfrac{5}{\sqrt{30}},85+2.04523\times\dfrac{5}{\sqrt{30}})

=(83.133,86.867)=(83.133, 86.867)

Therefore, based on the data provided, the 95% confidence interval for the population mean is 83.133<μ<86.867,83.133 < \mu < 86.867, which indicates that we are 95% confident that the true population mean μ\mu is contained by the interval (83.133,86.867).(83.133, 86.867).



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