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

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 Zo.os-1.64485362695147 to the value of to.os(n-1 d.f.). Give your answer to 4 decimal places. Hint: EXCEL: the comma nd is =TINV(two-tail probability, degrees of freedom), ie.-TINV(,n-1). OR R: gt(probability to the left, degrees of freedom), i.e·qt(14/2,n-1). t0.05 = 1.6509 ! 152-8083 (D) m b. Use the value of t0.05 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) ! 1522702 Hi c.Compare Zo.025-1.95996398454005 to the value of to.o25(n-1 d.f.). Give your answer to 4 decimal places. o.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) ! 152-3055

Explanation / Answer

(e) n=256 then df=n-1=256-1=255

variance=2*255=510

sd=sqrt(510)=22.58

for left 0.05 value=x-bar+z(0.05)*sd=255-1.6449*22.58=217.86

for right 0.05 value=x-bar+z(0.95)=255+1.6449*22.58=292.14

(f)chi-square(0.95,255)=219.03 and

chi-square(0.05,255)=293.25

(g) (1-alpha)% confidence interval for ={ sqrt(n-1)s2/chi-sq( alpha/2 ,n-1),sqrt(n-1)s2/chi-sq(1- alpha/2 ,n-1) }

90% confidence interval for ={ sqrt(n-1)s2/chi-sq( 0.1/2 ,n-1),sqrt(n-1)s2/chi-sq(1- 0.1/2 ,n-1) }

={sqrt(255*44*44/293.25),sqrt(255*44*44/219.03)}={41.03, 47.48}

(h) P(X>271)=P(z>0.7086)=1-P(z<0.7086)=1-0.7607=0.2393

(i)P(218<Y<265)=P(Y<265)-P(Y<218)=0.6711-0.0506=0.6205

(h)P(218<Y<265)=0.9549-0.3204=0.6345

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