Question

Assume that a sample is used to estimate a population mean H. Find the 99% confidence interval for a sample of size 372 with a mean of 48.4 and a standard deviation of 16.1. Enter your answer as a tri-linear inequality accurate to 3 decimal places. << Question Help: Message instructor Submit Question

Answer 1

1.The test statistic used will be "z " since it is large sample test so

Lower bound=Xm -Z1×S.E(Xm) where Xm is sample mean and S.E is standard deviation of sample mean, Z1 is critical value of z at 1% level of significance (obtained from z table) S.E(Xm)=$\sigma$/$\sqrt$n =16.1/$\sqrt$372 =0.8347

Putting the values:

=(48.4-2.576×0.8347) =48.4-2.1502= 46.2498=46.250

Upper bound = Xm+Z1×S.E(Xm)

=48.4+2.1502

=50.5502 =50.550

Ans: 46.250<mu<50.550

2. S.E (Xm)=$\sigma$/$\sqrt$n =9.3/$\sqrt$51 =1.3023

Z1=2.326 (from z table) At 98% level of significance.

Lower bound=Xm-Z1×S.E(Xm)= 26.3-2.326×1.3023 =23.2709= 23.270

Upper bound =Xm +z1×s.e(Xm) =26.3+2.326×1.3023

=29.3291 =29.329

Ans: 23.170<mu<29.329

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