Question

problem 11

10. Confidence Intervals Repeat Problem 9 using a confidence level of 95 % and sample size of 1,000. (Sections 6.1 & 6.2) Mean repair cost $143, s = $35 o Margin of error, E. a5 Confidence Interval: SLOH 45.as Mean repair cost $143 , a 35. 140.83 Margin of error, E. Confidence Interval: <με. LISHI O SBs Which confidence interval is wider?Ost1u 3 11 Confidence Interval Comparison when sample size is 32 and 1.000 Use the information in problems 9 and 10. Compare the width differential when the sample size is 1,000 with the width differential when the sample size is 35 (in problem 9)? a. b. What can you conclude about the Normal and t-distributions as the sample sizes increase? 6

Answer 1

11)

Mean = 143

S.d = 35

When n = 35

As the population s.d is unknown we will use t distribution to estimate the interval

Degrees of freedom is = n-1 = 34

Critical value t from table for 34 degrees of freedom and 95% confidence level is = 2.03

Margin of error = t*s.d/√n = 2.03*35/√35 = 12.02

Required interval is

(Mean - MOE, Mean + MOE)

( 130.98, 155.02 )

Width = 155.02 - 130.98 = 24.04

When N =1000

Degrees of freedom is = n-1 = 999

For 999 df and 95% confidence level critical value t is = 1.96

Margin of error = 1.96*35/√1000 = 2.17

Interval is

(140.83, 145.17)

Width = 145.17 - 140.83 = 4.34

B)

As the sample size increases, width of the interval decreases

And also as the sample size increases, t distribution approaches to normal distribution