Question

A manufacturer of home products is designing a new wall mount for large flat-panel televisions. Since some flat-panel televisions weigh up to 100 pounds, the manufacturer wants to make sure that the mean weight of failure for the wall mounts is greater than 180 pounds. Dayton is a researcher for the manufacturer who is conducting the test. He randomly selects 35 wall mounts and tests them by using a machine that pulls down on the mounts until each one fails. The results are shown in the data set provided. Based on the research completed during the development phase, Dayton assumes that the population standard deviation is 6.9 pounds.

Use Excel to test whether the mean failure weight is greater than 180 pounds. Calculate the test statistic, z, rounding to two decimal places, and the p-value, rounding to three decimal places.

Failure Weight (pounds)
177
171
193
186
188
175
191
190

z=

p-value=
193
184
186
190
198
201
186
195
181
189
193
195
178
194
185
187
180
192
196
194
182
183
179
188
187
181
197

Answer 1

Solution: Here concept of one sample z-test for mean is used.

Solution- alternative hypotheses follows: The rull and H : u = 180 u > 180 Ha: z- test statisti hypothesis shall To test the Use which will follows : be Ous 6-/55 where, is sample mian M. in specified valve of u nder H. is population standard deviation is sample size We have. 35, + 137) Using excel ΣΧ - 6565 1=1 6565 187 5714 and 6.3 M. = 180 187.5714 -180 6.49 6.9//35 The valve test statistic is is right- toiled test. The p-valve (right-taibd) test Our is geven by, P(z> valve of the test statistie) P- valve = P(z> 6.43)

P- valve O.000 %3D follows: make decision rule We P-valve > significane level (a) , at giren significane level · significana level (a) , then then we fail If to reject Ho we reject If P-valve H. at giren significance level. is not given- In the question significance level (a) Gener ally significance lared og oof or X = 0.05 X = 0.01 ie. Genurally or used. ie. d= 0.01 is sigmi ficance lered 0.01 Let our significance level (a) = 0.01 P-valve 0.000 %3D L0.000 < 0.01) P-valve < significance level og 0.01, reject thuufore Since, at 0.01 significance level. H. shall significance lerel of ools , there is support the claim Conclusion- At eviden ce sufficient to for the wall monts weight of is greater than failure that mean 180 ponds

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