The width of a sheet of standard size copier paper should be = 216 mm (i.e., 8.5

Question

The width of a sheet of standard size copier paper should be = 216 mm (i.e., 8.5 inches). There is variation in paper width due to the nature of the production process, so the width of a sheet of paper is a random variable. Samples are taken from the production process and the mean width is calculated. If the paper is too narrow, the pages might not be well-centered in the sheet feeder. If the pages are too wide, sheets could jam in the feeder or paper trays. Either violation poses a quality problem, so the manufacturer might choose a two-tailed test. Download data file paperwidth.xlsx 1. Find the sample mean of the data with 95% confidence interval. 2. Is the width mean of a sheet of standard size copier paper 216 mm? Do the hypothesis test using the 5% level of significance using the following steps : Step 1: State the Hypotheses Step 2: Specify the Decision Rule Step 3: Calculate the Test Statistic Step 4: Make the Decision Calculate p-value also and give interpretations. Width of Sheets of Paper in Millimeters (n = 50) Width (mm) 216.025 216.021 215.978 216.064 216.02 216.012 216.033 215.981 216.013 216.026 215.999 216 216.019 216.019 215.972 215.993 215.998 215.955 215.962 216.012 216.037 215.985 216.035 215.996 216.018 216.032 216.006 216.003 215.986 215.99 216.002 216.039 216.039 215.983 215.998 216.002 215.994 215.997 216.007 216.036 216.032 216.031 216.011 216.004 216.011 215.999 215.976 216 216.018 215.981

Explanation / Answer

Given that,
population mean(u)=216
sample data:
( 216.025 216.021 215.978 216.064 216.02 216.012 216.033 215.981 216.013 216.026 215.999 216 216.019 216.019 215.972 215.993 215.998 215.955 215.962 216.012 216.037 215.985 216.035 215.996 216.018 216.032 216.006 216.003 215.986 215.99 216.002 216.039 216.039 215.983 215.998 216.002 215.994 215.997 216.007 216.036 216.032 216.031 216.011 216.004 216.011 215.999 215.976 216 216.018 215.981 )
From the above sample data
caluclated sample mean, x =216.007
caluclatedd sample standard deviation, s =0.02212
number (n)=50
null, Ho: =216
alternate, H1: !=216
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.01
since our test is two-tailed
reject Ho, if to < -2.01 OR if to > 2.01
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =216.007-216/(0.02212/sqrt(50))
to =2.2377
| to | =2.2377
critical value
the value of |t | with n-1 = 49 d.f is 2.01
we got |to| =2.2377 & | t | =2.01
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.2377 ) = 0.0298
hence value of p0.05 > 0.0298,here we reject Ho
ANSWERS
---------------
null, Ho: =216
alternate, H1: !=216
test statistic: 2.2377
critical value: -2.01 , 2.01
decision: reject Ho
p-value: 0.0298

Hence we have evidence to say that the width mean of a sheet of standard size copier paper 216 mm

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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