Please help me answer this question Homework: Chapter 12 Homework Save Score: 0.

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Please help me answer this question

Homework: Chapter 12 Homework Save Score: 0.17 of 1 pt 2 of 2 (1 complete) Hw Score: 58.33%, 1.17 of 2 pts 12.2.23-T Question Help A company manufactures and ships four different products. Breakage cost is expensive, and the company would like to select a mode of delivery (rail, plane, or truck) that reduces the amount of breakage. The company decided to examine the dollar amount of breakage incurred by the three modes of transportation. In order to control for fragility differences due to type of product, the company randomly assigns each product to each carrier and monitors the dollar breakage that occurs over the course of 100 shipments. The dollar breakage per shipment (to the nearest dollar) is shown in the accompanying table. Complete parts a and b. Click the icon to view the data. a. Was the company correct in its decision to block for type of product? Conduct the appropriate hypothesis test using a level of significance of 0.05. Determine the null and alternative hypotheses for this test. Choose the correct answer below. O B. Ho: Not all means are equal (blocking is effective) HA: Not all means are equal (blocking is effective) ° C. Ho-Frail-Uplane-Pruck O D. Ho : prod 1 brodz" prod3" prod4 (blocking is effective) HA: At least two populations have different means HO: At least two populations have different means HA Ntl means are equal 0 E, HA: All means are unequal (blocking is effective) What is the F-test statistic for this test? F-(Round to two decimal places as needed.)

Explanation / Answer

applying two way ANOVA without replication:

a)

option A is correct

F = 0.71

F0.10 =3.289 (for 3 and 6 degree of freedom)

the test statistic is less then the critical vlaue , so fail to reject the null hypothesis. there is not sufficent evidence to ......

b)

option E is correct

F =0.20

F0.10 =3.463 (for 2 and 6 degree of freedom)

the test statistic is less then the critical vlaue , so fail to reject the null hypothesis. there is not sufficent evidence to ......

SUMMARY Count Sum Average Variance Product 1 3 24559 8186.333 3523985 Product 2 3 26890 8963.333 781932.3 Product 3 3 22997 7665.667 626520.3 Product 4 3 21356 7118.667 3426874 rail 4 31193 7798.25 1406370 plane 4 33608 8402 2378161 truck 4 31001 7750.25 3286693 ANOVA Source of Variation SS df MS F P-value Rows 5550508 3 1850169 0.7087 0.5811 Columns 1055462 2 527730.8 0.2022 0.8223 Error 15663163 6 2610527 Total 22269133 11

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