Trough Widths (in inches) 8.3158.315 8.3448.344 8.3098.309 8.3838.383 8.3518.351
ID: 3312730 • Letter: T
Question
Trough Widths (in inches)
8.3158.315
8.3448.344
8.3098.309
8.3838.383
8.3518.351
8.4068.406
8.3478.347
8.3758.375
8.4878.487
8.4258.425
8.4838.483
8.3768.376
8.4798.479
8.4118.411
8.4178.417
8.4258.425
8.3798.379
8.4658.465
8.5018.501
8.4528.452
8.4398.439
8.4118.411
8.4848.484
8.4138.413
8.4838.483
8.4088.408
8.4768.476
8.4248.424
8.4578.457
8.4658.465
8.4528.452
8.4518.451
8.4258.425
8.4628.462
8.4088.408
8.4198.419
8.4138.413
8.4068.406
8.3178.317
8.4258.425
8.4028.402
8.4428.442
8.4028.402
8.4378.437
8.4098.409
8.4228.422
8.4268.426
8.5028.502
8.4148.414
8.4178.417
Determine the test statistic, p-value & State the concern
Trough Widths (in inches)
8.3158.315
8.3448.344
8.3098.309
8.3838.383
8.3518.351
8.4068.406
8.3478.347
8.3758.375
8.4878.487
8.4258.425
8.4838.483
8.3768.376
8.4798.479
8.4118.411
8.4178.417
8.4258.425
8.3798.379
8.4658.465
8.5018.501
8.4528.452
8.4398.439
8.4118.411
8.4848.484
8.4138.413
8.4838.483
8.4088.408
8.4768.476
8.4248.424
8.4578.457
8.4658.465
8.4528.452
8.4518.451
8.4258.425
8.4628.462
8.4088.408
8.4198.419
8.4138.413
8.4068.406
8.3178.317
8.4258.425
8.4028.402
8.4428.442
8.4028.402
8.4378.437
8.4098.409
8.4228.422
8.4268.426
8.5028.502
8.4148.414
8.4178.417
9.2.32-T Question Help A company that manufactures steel troughs requires that the width of the trough be between 8.31 inches and 8.61 inches. The data in the accompanying table contains the widths of the troughs, in inches, for a sample of n-50.Complete parts (a) through (d). Click the icon to view the data table. a. At the 0.01 level of significance, is there evidence that the mean width of the troughs is different from 8.46 inches? State the null and alternative hypotheses. Type integers or decimals.) Enter your answer in the edit fields and then click Check Answer. 6 Pemaining Clear All Check Answer 5 7 8 9Explanation / Answer
Given that,
population mean(u)=8.46
sample mean, x =8.42082
standard deviation, s =0.0466
number (n)=50
null, Ho: =8.46
alternate, H1: !=8.46
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.68
since our test is two-tailed
reject Ho, if to < -2.68 OR if to > 2.68
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.42082-8.46/(0.0466/sqrt(50))
to =-5.9452
| to | =5.9452
critical value
the value of |t | with n-1 = 49 d.f is 2.68
we got |to| =5.9452 & | t | =2.68
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -5.9452 ) = 0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =8.46
alternate, H1: !=8.46
test statistic: -5.9452
critical value: -2.68 , 2.68
decision: reject Ho
p-value: 0
mean value is dffrent from 8.46
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