# q1) The 95% confidence interva

q1) The 95% confidence interval for the mean, calculated from asample is 1.412011 ≤ μ ≤ 2.587989. Determine the sample meanX(this one x dash) =

q2) Assuming that the data is normally distributed with thepopulation standard deviation =1.5, determine the size of thesample n =

confidence interval is                 lower limit =    1.412011             upper limit=   2.587989             sample mean = (lower limit+upper limit)/2= (  1.412011   +   2.587989   ) / 2=   2

2)

margin of error = (upper limit-lower limit)/2= (  2.587989   –   1.412011   ) / 2=   0.587989alpha =   1-CL =   5%                Z value =    Zα/2 =    1.960   [excelformula =normsinv(α/2)]                                     Sample Size,n = (Z * σ / E )² = (   1.960  *   1.5   /   0.587989   )² =   25.000                                                So,Sample Size needed=       25

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