Question

Feit Electric manufactures 60-watt equivalent Compact Fluorescent Bulbs that it guarantees will last at least 10,000 hours. The Consumer Testing Company tested the claim of the manufacturer by purchasing a random sample of 60-watt equivalent CFBs manufactured by Feit Electric. The Consumer Testing Company tested 36 CFBs. The results of the test are found in Chapter_10_Project_HT_CFB.xlsx. The Consumer Testing Company has no other data on CFBs from the manufacturer, so must rely for its test on the sample of 36 bulbs. Use Proposed R Code for Part Two 2. and the Excel file Chapter_10_Project_HT_CFB.xlsx. Calculation should be completed in RStudio. For the null hypothesis, assume that the Compact Fluorescent Bulbs satisfy the guarantee of Feit Electric.

(a) State the null and the alternative hypothesis. H0 : µ = Ha : µ [ ≠ or > or < ]

(b) Should the test be a left-tail, right-tail or two- tail test?

(c) Calculate t using the R code for variable tcfb given below. [Proposed R Code for Part Two 2.]

(d) Find the rejection region (critical value) for α = 0.01. What is the decision? [Compare tcfb to tcrit01.] Is the null hypothesis rejected or do the CFBs satisfy the guarantee?

(e) Find the rejection region (critical value) for α = 0.05. What is the decision? Compare tcfb to tcrit05.] Is the null hypothesis rejected or do the CFBs satisfy the guarantee?

DATA. 8800

9155

13001

10250

100002

11413

8234

10402

10016

8015

6110

11005

8550

11555

9254

6991

12006

10420

8302

8151

10980

8694

10186

10003

8814

11445

6277

8632

7265

10584

9397

11987

8502

7556

10380

8582

Answer #1

(a) the null and the alternative hypothesis. H0 : µ =10000

Ha : µ <10000

(b) the test should be a left-tail,

using R

we have

>
x=c(8800,9155,13001,10250,100002,11413,8234,10402,10016,8015,6110,11005,8550,11555,9254,6991,12006,10420,8302,8151,10980,8694,10186.10003,8814,11445,6277,8632,7265,10584,9397,11987,8502,7556,10380,8582)

> t.test(x,alternative =c("less") ,mu=10000 )

One Sample t-test

data: x

t = 0.7784, df = 34, p-value = 0.7791

alternative hypothesis: true mean is less than 10000

95 percent confidence interval:

-Inf 16427.39

sample estimates:

mean of x

12026.09

(c) t = 0.7784

(d) the rejection region (critical value) for α = 0.01 is if t < -2.44 the null hypothesis is not rejected

(e) the rejection region (critical value) for α = 0.05 is t < -1.69 Is the null hypothesis is not rejected

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