A random sample of n=400 students is selected from a population of N=4000 students to estimate the average weight of the students. The sample mean and sample variance are found to be x=140lbands2=225. Find the 95%(z=2) confident interval.

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A random sample of n=400 students is selected from a population of N=4000 students to estimate the average weight of the students. The sample mean and sample variance are found to be x=140lbands2=225. Find the 95%(z=2) confident interval.

liannemdh · Accepted Answer

Step 1  The (1−α)100% confidence interval formula for the population mean when population standard deviation is not known, is defined as follows:  CI=x―±tα2,n−1(sn).  tα2,n−1 is the critical value of the t-distribution with degrees of freedom of n−1 above which, 100(α2)%orα2 proportion of the observations lie, and below which, 100(1−α+α2=100(1−α2)%or(1−α2) proportion of the observation lie, x― is the sample mean, s is the sample standard deiation, and n is the sample size.  Step 2  The sample mean is x―=140lb, and the sample standard deviation is s=15lb(=√225).  The sample size is n=400. The degrees of freedom is 399(=400–1).  The confidence level is 0.95. Hence, the level of significance is 1–0.95=0.05.  Using Excel formula: =T.INV.2T(0.05,399), the critical value is, tα2=t0.025≈1.96.  Thus,  CI=x―±tα2,n−1(sn)  =140±(1.96)(15400)  =140±(1.96)(0.75)  =140±1.47  =(138.53,141.47)  Thus, the required 95% confidence interval is (138.53 lb, 141.47 lb).

A random sample of displaystyle{n}={400} students is selected from a population of displaystyle{N}={4000} students to estimate the average weight of t

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