A random sample of size n 256 from a N(, 2) distribution gives the sample mean x

Question

A random sample of size n 256 from a N(, 2) distribution gives the sample mean x = 740 and the sample standard deviation s = 44. If the number of degrees of freedom (d.f.) is large, the values of ta/2 can be approximated by Za/2. a. Compare Z0.05= 1.64485362695 147 to the value of t0.05(n-1 d.). Give your answer to 4 decimal places. Hint: EXCEL: the command is-T1NV(two-tail probability, degrees of freedom), i.e.-T1NV(,n-1). OR R: qt(probability to the left, degrees of freedom), i.e. qt(1-a/2,n-1) t0.05 = 1.6509 ! 152-8083 b. Use the value of to.05 in part (a) to construct a 90% (two-sided) confidence interval for . Round your answer to 2 decimal places 90% CI for = ( 735.46 , 744.54 ) ! 152-27020 c. Compare Zo.025-1.95996398454005 to the value of to.o2s(n-1 d.f.). Give your answer to 4 decimal places. t0.025 = 1.9693 ! 152-7203 d. Use the value of t0.025 in part (c) to construct a 95% (two-sided) confidence interval for . Round your answer to 2 decimal places 95% CI for = ( 734.58 , 745.42 ) ! m 152-30550

Explanation / Answer

Solution:

a)We are give n = 256 , xbar = 740 , s =44

and Z0.05 = 1.6448

So , t0.05 = 1.6509 Using excel , =TINV(1-(0.05/2),255)

We can see that , Z0.05 <   t0.05 .

b)Now we have to find 90% confidence interval for using t0.05

M = 740
t = 1.65
sM = (442/256) = 2.75

= M ± t(sM)
= 740 ± 1.65*2.75
= 740 ± 4.54

90% CI [735.46, 744.54].

c) Z0.025 = 1.9596 t0.025 = 1.9693

We can see that , Z0.05 <   t0.05 .

d) M = 740
t = 1.97
sM = (442/256) = 2.75

= M ± t(sM)
= 740 ± 1.97*2.75
= 740 ± 5.42

95% CI [734.58, 745.42]

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