stat analysis homework and excel lab
farhm9
lab_3_tutorial.xlsSheet1
| Tutorial | ||||||
| DS 123 Lab #3 | ||||||
| z-score on sample mean, s known, measurements | ||||||
| Question 1. | Suppose the federal government proposes to give a tax break to automakers | |||||
| producing midsize cars that get a mean mileage exceeding 31 mpg. Let m be the | ||||||
| midsize car's mean mileage and assume that s equals 0.8. For a sample of 49 cars | ||||||
| produced by one automaker, the average sample mileage was 31.5531 mpg. Use the | ||||||
| 1% level of significance to test whether the automaker should be awarded the tax break. | ||||||
| z-Statistic | p-value | |||||
| Hypotheses: | ||||||
| Rejection Region: | a = 0.01 | a = 0.01 | ||||
| z-coeff = | 2.326347874 | Rej H0 if p-value < 0.01 | ||||
| Rej H0 if z > 2.326 | ||||||
| Test Statistic: | z = | 4.839625 | p-value = | 0.0000006504 | ||
| Conclusion: | ||||||
| Since the test statistic of z = 4.84 falls | Since the test statistic of | |||||
| well beyond the critical value of z = 2.326, | p-value = 0.00000065 falls below | |||||
| we reject H0 with at least 99% | alpha = 0.01, we reject H0 | |||||
| confidence. | with at least 99% confidence. | |||||
| Implication: | ||||||
| There is enough evidence to conclude | There is enough evidence to | |||||
| that the automaker should be awarded | conclude that the autom,aker should | |||||
| the tax break. | be awarded the tax break. | |||||
| z-score on proportion, counts | ||||||
| Question 2. | Problem 9.39, p. 365. Use a = 0.05 level of significance. | |||||
| # of people recalling the commercial w/o prompting | 46 | |||||
| Total # people sampled | 200 | |||||
| Sample proportion = | 0.23 | |||||
| t-Statistic | p-value | |||||
| Hypotheses: | ||||||
| Rejection Region: | a = 0.05 | a = 0.05 | ||||
| z-coeff = | 1.644853627 | Rej H0 if p-value < 0.05 | ||||
| Rej H0 if z > 1.645 | ||||||
| Test Statistic: | z = | 1.8405254346 | p-value = | 0.0328455668 | ||
| Conclusion: | ||||||
| Since the test statistic of z = 1.84 falls | Since the test statistic of | |||||
| well beyond the upper critical value of | p-value = 0.0328 falls | |||||
| z = 1.645, we reject H0 with | below a = 0.05, we reject H0 | |||||
| at least 95% confidence. | with at least 95% confidence. | |||||
| Implication: | ||||||
| There is enough evidence to conclude | There is enough evidence to | |||||
| that the mouthwash commercial | conclude that the mouthwash | |||||
| is successful. | commercial is successful. |
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