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Quality Associates

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1. State the null and the alternative hypotheses in this scenario (4 pts): A new sales force bonus plan is developed in an attempt to increase sales.
Null hypothesis: New bonus plan does not increase sales.
Alternative hypothesis:--New bonus plan increases sales.

2. For this question refer to Case Problem 1 (Quality Associates Inc.) on page 410 of your text (30 pts)
Sample 1 1) H0: μ = 12
Ha: μ ≠ 12 2) α = .01, but for two-tail test will = .005 3) Z = (x-bar – μ) / (σ/√n) 4) Z Critical value at .005 = 2.575 5) Z = (11.9587 – 12) / (.21/√30) = -1.077187
The observed value lies outside the rejection region, so we fail to reject H0. 6) P –value is between .2814 for a two-tailed test
Sample 2 1) H0: μ = 12
Ha: μ ≠ 12 2) α = .01, but for two-tail test will = .005 3) Z = (x-bar – μ) / (σ/√n) 4) Z Critical value at .005 = 2.575 5) Z = (12.0287 – 12) / (.21/√30) = .74855
The observed value lies outside the rejection region, so we fail to reject H0. 6) P-value is 0.4541 for a two-tailed test
Sample 3 1) H0: μ = 12
Ha: μ ≠ 12 2) α = .01, but for two-tail test will = .005 3) Z = (x-bar – μ) / (σ/√n) 4) Z Critical value at .005 = 2.575 5) Z = (11.889 – 12) / (.21/√30) = - 2.895
The observed value lies inside the rejection region, so we reject H0. 6) P-Value is .0038 for a two-tailed test

Sample 4 1) H0: μ = 12
Ha: μ ≠ 12 2) α = .01, but for two-tail test will = .005 3) Z = (x-bar – μ) / (σ/√n) 4) Z Critical value at .005 = 2.575 5) Z = (12.081 – 12) / (.21/√30) = 2.11264
The observed value lies outside the rejection region, so we fail to reject H0. 6) P-value is .034 for a two-tailed test
2) Sample 1 Standard Deviation = 0.22035603
Sample 2 Standard Deviation = 0.22035603
Sample 3 Standard Deviation = 0.207170594
Sample 4 Standard Deviation = 0.206108999
Yes it appears

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