Question

11.1

The number of miles of useful tread wear (in 1000s) was determined for tires of each of five different makes of subcompact car (factor A, with 5) in combination with each of four different brands of radial tires factor B, with -4 resulting in = 20 observations. The values SSA 30 9 SSB 44.3, and 592 were then computed. Assume that an additive mode appropriate. (a) Test Ho: αι = a 2 = α3 α4-as-0 (no differences in true average tire lifetime due to makes of cars) versus Ha. at least one α decimal places.) Calculate the test statistic. (Round your answer to two decimal places.) 0 using a level 0.05 test. (Round your answers to two What can be said about the P-value for the test? P-value>0.100 0.050 P-value<0.100 0.010 < P-value <0.050 0.001 < P-value <0.010 р.value 0.00 1 State the conclusion in the problem context. Reject Ho The data suggests that there are differences in true average tire lifetimes due to makes of cars. Fail to reject Ho. The data suggests that there are differences in true average tire lifetimes due to makes of cars. Fail to reject Ho- The data does not suggest that there are differences in true average tire lifetimes due to makes of cars. Reject Ho. The data does not suggest that there are differences in true average tire lifetimes due to makes of cars. (b) Test Ho: / 1-2 = 3 = 4-0 (no differences in true average tire lifetime due to brands of tires versus Hai at least one 0 using a level 0.05 test. (Round your answers to two decimal places.) Calculate the test statistic. (Round your answer to two decimal places.)

Answer 1

Part (a)

i) Test statistic

= SSA/SSE

= 30.9/59.2

= 0.52 Answer 1

ii) p-value

= P(F_{4, 12} > 0.5220)

= 0.7216 First Option Answer 2

iii) Since p-value is very high (greater than any conceivable significance level), the null hypothesis of no difference in effect is accepted.

Hence Third Option Answer 3

Part (b)

Test statistic

= SSB/SSE

= 44.3/59.2

= 0.75 Answer 4

DONE

[Going beyond, (b) Part’s natural continuation

p-value

= P(F_{3, 12} > 0.7483)

= 0.5439

Conclusion: Since p-value is very high (greater than any conceivable significance level), the null hypothesis of no difference in effect is accepted.

Complete