Question

Hello experts,

I need help answering the following question and its sub-questions please. I use MINITAB for my data. Please show your work to make it easier for understanding! will upvote!

Q-1) 7 marks Confidence interval for μ when σ 1s known. The duration of baseball games in the major league in the 2010 and 2011 scasons is uniformly distributed between 2.76 and 3.22 hours inclusive. We would like to simulate 50 samples of 30 gane duration time and construct a 90% confidence interval for the mean duration of 30 games for each of the samples: MTBrandom 50 c5-34 SUBC> Uniform 2.76 3.22. MTB >mean c5-c3, c2 MTB let c3 c2.645 0.1328/sqrt (30) MTB let c4c2 1.645 0.1328/sqrt (30) MTB > let cl-(c3< 2.99 and c4 2.99) MTB > tally cl Next page STAT-2507 Introduction to Statistical Modelling Winter 2019 (a) (2 pts) Attach your worksheet from cl-c4. (b) (2 pts) What percentage of these 50 constructed 90% confidence intervals did actually contain the population mean μ = 2.997 (c) (1 pt) Based on the interpretation of a 90% confidence interval, is this the number expected? (d) (2 pts) If you repeat the simulation (30 duration time in one sample and 50 samples in total) one more time, do you expect the same answer in part (c)? Justify your answer.

Answer 1

Means	c3	c4	c1
3.010848	2.970963	3.050732	TRUE
3.008853	2.968968	3.048737	TRUE
2.994559	2.954674	3.034443	TRUE
2.983877	2.943992	3.023761	TRUE
3.029578	2.989694	3.069463	TRUE
2.948921	2.909036	2.988805	FALSE
2.97329	2.933406	3.013175	TRUE
2.983662	2.943778	3.023546	TRUE
2.976328	2.936444	3.016213	TRUE
3.039224	2.99934	3.079109	FALSE
2.985974	2.94609	3.025858	TRUE
2.998887	2.959002	3.038771	TRUE
2.975275	2.935391	3.01516	TRUE
3.027068	2.987184	3.066952	TRUE
2.952829	2.912944	2.992713	TRUE
2.98367	2.943786	3.023555	TRUE
3.034941	2.995056	3.074825	FALSE
3.011675	2.971791	3.05156	TRUE
3.000894	2.961009	3.040778	TRUE
2.995119	2.955235	3.035004	TRUE
3.034452	2.994567	3.074336	FALSE
3.017926	2.978041	3.05781	TRUE
2.995006	2.955122	3.034891	TRUE
2.989743	2.949859	3.029628	TRUE
2.955542	2.915658	2.995427	TRUE
2.989964	2.950079	3.029848	TRUE
3.024626	2.984741	3.06451	TRUE
3.006223	2.966339	3.046108	TRUE
3.038681	2.998796	3.078565	FALSE
2.957377	2.917493	2.997261	TRUE
2.965837	2.925953	3.005722	TRUE
2.973263	2.933379	3.013147	TRUE
3.001434	2.96155	3.041318	TRUE
2.973601	2.933716	3.013485	TRUE
2.952509	2.912624	2.992393	TRUE
2.959724	2.91984	2.999609	TRUE
3.006242	2.966358	3.046126	TRUE
3.006776	2.966892	3.046661	TRUE
2.976261	2.936377	3.016145	TRUE
2.980785	2.940901	3.020669	TRUE
3.000447	2.960563	3.040332	TRUE
3.006007	2.966123	3.045892	TRUE
2.976691	2.936807	3.016576	TRUE
2.96029	2.920406	3.000174	TRUE
3.030931	2.991047	3.070816	FALSE
2.987327	2.947442	3.027211	TRUE
2.99311	2.953225	3.032994	TRUE
2.985773	2.945889	3.025658	TRUE
2.969324	2.92944	3.009209	TRUE
2.974005	2.934121	3.01389	TRUE

> Data<-data.frame(replicate(30,runif(50,min=2.76,max=3.22)))
> Data$Means=rowMeans(Data[,-1])
> Data$c3=Data$Means-1.645*0.1328/sqrt(30)
> Data$c4=Data$Means+1.645*0.1328/sqrt(30)
> Data$c1=(Data$c3<2.99 & Data$c4 >2.99)
> summary(Data$c1)
Mode FALSE TRUE
logical 6 44
> View(Data)

P=44/50=0.88

88% of the samples contain the mean.

Based on the interpretation of a 90% confidence interval, on average 90% of confidence intervals will contain actual mean.

d) We take 50 samples of size 30 each.  

> Data<-data.frame(replicate(50,runif(30,min=2.76,max=3.22)))
> Data$Means=rowMeans(Data[,-1])
> Data$c3=Data$Means-1.645*0.1328/sqrt(50)
> Data$c4=Data$Means+1.645*0.1328/sqrt(50)
> Data$c1=(Data$c3<2.99 & Data$c4 >2.99)
> write.csv(Data,"Data.csv")
> summary(Data$c1)
Mode FALSE TRUE
logical 3 27

p=27/30=  0.9

The results will be around 90%. The true proportion is around 90% not necessarily 90% every time.