Question

The XYZ company manufactures a new kind of light bulb based on the latest LED technology. The (population) mean life of a XYZ bulb is 9 years with a population standard deviation of 1.2 years. The life of a XYZ bulb is normally distributed.

What proportion of XYZ bulbs will last between 9 and 12 years?

A drug company claims that its new painkiller has exactly 6 mg. of codeine. You test the claim at a significance level (alpha) of .05. You randomly sample 100 pills made by the company and find the sample mean to be 5 milligrams of codeine with a sample standard deviation of .75 milligrams.

What is your conclusion about the claim?

Select one:

reject

accept

Select one:

0.5862

0.1125

0.962

0.4938

The ABC Company manufactures printers. You have been asked to construct a confidence interval (CIE) that can be used to estimate the population mean life. You take a random sample of 100 scanners and find the sample mean life to be 10 years with a sample standard deviation of 2 years. Construct the following confidence interval estimators (CIE). You may use the values from the Z distribution since n is quite large.

Construct an 82% CIE

Select one:

10 ± .357

10 ± .268

10 ± .568

12 ± .268

Answer 1

1)proportion of XYZ bulbs will last between 9 and 12 years

probability =P(9<X<12)=P((9-9)/1.2)<Z<(12-9)/1.2)=P(0<Z<2.5)=0.9938-0.5=0.4938

2)

null hypothesis: HO: μ		=	6
Alternate Hypothesis: Ha: μ		≠	6
0.05 level with two tail test and n-1= 99 df, critical t=				1.984
Decision rule :reject Ho if absolute value of test statistic|t|>1.984
	population mean μ=	6
	sample mean 'x̄=	5.000
	sample size   n=	100.00
	sample std deviation s=	0.750
	std error 'sx=s/√n=	0.075
	test stat t ='(x-μ)*√n/sx=	-13.333

reject ho , since test statistic falls in rejection region

3)

sample mean 'x̄=	10.000
sample size    n=	100.00
std deviation σ=	2.000
std error ='σx=σ/√n=	0.2000

for 82 % CI value of z=		1.3408
margin of error E=z*std error =		0.268

82% CIE =10 -/+ 0.268