MTH128- Stats (Online ist 8 weekSI-Setion 9u6 Save Homework: Chapter5 Score: 0 o

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MTH128- Stats (Online ist 8 weekSI-Setion 9u6 Save Homework: Chapter5 Score: 0 of 1pt 5.4.35 35 of 36 (27 complete) Hw Score: 58.56%, 21.08 of 36 pts -Question Help * A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 134 ounces and a standard deviation of 0.30 ounce You randomly select 50 cans and carefully measure the contents. The sample mean of the cans is 133.9 ounces. reasoning Does the machine need to be reset? Explain your |Within the range of a usual 1, it is that you would have randomly sampled 50 cans with a mean equal to 133.9 ounces, because it of the mean of the sample means. event, namely within Click to select your answer(s) and then click Check Answer Clear A All parts showing (MTH128- Stats (Online 1st 8 Weeks)-Section 508) is based on Larson Elementary Statistics Picturing the World, 6e

Explanation / Answer

let X1,X2,...,Xn denotes the contents of cans.

assumption is X1,X2,...,Xn follows N(u,sigma2) independently.

now sigma is knwon to be 0.30 ounces

the machine can dispense paint at a mean of 134 ounces.

we need to test whether the machine needs to be reset.

so H0: u=134 vs H1:u>134

to test this we have a random sample of n=50 cans with mean contents=Xbar=133.9 ounces.

so the test statistic is given as T=(Xbar-134)*sqrt(n)/sigma which under H0 follows a N(0,1) distribution.

assuming the level of significance to be 0.05, and since the alternate hypothesis is right sided the H0 is rejected iff

t>z0.05 where t is the observed value of T and z0.05 is the upper 0.05 point of a N(0,1) distribution.

now t=(133.9-134)*sqrt(50)/0.30=-2.357

and z0.05=1.64

so t<z0.05

hence at 5% level of significance H0 is not rejected.

the 99% confidence interval is now calculated.

now Xbar being the mean of X1,...,Xn

so Xbar~N(u,0.302/n)   

or P[|Xbar-u|*sqrt(n)/0.30<z0.005]=0.99

or, P[Xbar-z0.005*0.30/sqrt(n)<u<Xbar+z0.005*0.30/sqrt(n)]=0.99

hence 95% confidence interval of u wth Xbar=133.9 and n=50 is

[133.9-2.58*0.30/sqrt(50),133.9+2.58*0.30/sqrt(50)]=[133.7905,134.0095] which contains 134.

hence no, it is possible that you would have randomly sampled 50 cans with a mean equal to 133.9 ounces, because it falls within the range of a usual event,namely within 99% confidendenc interval of the mean of the sample means

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