The manufacturer of a toner cartridge claims the mean number of printouts is 10,

Question

The manufacturer of a toner cartridge claims the mean number of printouts is 10,000 for each cartridge. A consumer advocate believes the actual mean number of printouts is lower. He selects a random sample of 14 such cartridges and obtains the following number of printouts (the data is normally distributed):

9,600 --- 10,300----9,000---10,750 ----9,490---9,080----9,655

9,520--10,070----9,999-----10,470-----8,920----9,964----10,330

a. Caculate the 95% confidence interval

b.Test the hypothesis Ho: µ = 10,000 and H1: µ < 10,000. Is there enough evidence to support the advocate’s claim at the = 0.05 level of significance? Use the classical and p value approach

THANKS FOR THE HELP , ITS SUPPOUSE TO BE A TABLE BUT FOR SOME REASON I CAN'T !

Explanation / Answer

A)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    9796.285714          
t(alpha/2) = critical t for the confidence interval =    2.160368656          
s = sample standard deviation =    567.7098973          
n = sample size =    14          
df = n - 1 =    13          
Thus,              
              
Lower bound =    9468.499778          
Upper bound =    10124.07165          
              
Thus, the confidence interval is              
              
(   9468.499778   ,   10124.07165   ) [ANSWER]

******************************

b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   10000  
Ha:    u   <   10000  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    13          
tcrit =    -   1.770933396      
              
Getting the test statistic, as              
              
X = sample mean =    9796.285714          
uo = hypothesized mean =    10000          
n = sample size =    14          
s = standard deviation =    567.7098973          
              
Thus, t = (X - uo) * sqrt(n) / s =    -1.342638319          
              
Also, the p value is              
              
p =    0.101182159          
              
As t > 1.771, and P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

There is no significant evidence that the actual mean number of printouts is lower that 10000. [CONCLUSION]

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.