The Historical Output B Employees Is A Mean Of 120 Units Per Hour W
The assignment involves evaluating several hypotheses about population means based on sample data, using significance testing at the 5% level. The scenarios include testing whether a new employee's output differs significantly from historical data, determining if a new battery's lifetime deviates from expected, assessing whether boys in a children's home are underfed based on weight data, evaluating a principal's claim about student IQ, and analyzing whether a class's vocabulary score is representative of typical performance. Each case requires performing a hypothesis test—either a z-test for known population standard deviations or a t-test when population variance is unknown—and interpreting the results within the context of significance levels, typically α = 0.05.
Paper For Above instruction
The process of hypothesis testing is fundamental in statistics to determine if observed data significantly deviate from what is expected under a given assumption or null hypothesis. Each scenario here involves formulating null hypotheses (H■) that assume no difference or effect, and alternative hypotheses (H■) that suggest a significant difference exists. By calculating test statistics and comparing them with critical values at a specified significance level (α = 0.05), conclusions can be drawn regarding the validity of these hypotheses.
In the first scenario, a new employee’s output is compared against the historical mean of 120 units/hour with a known standard deviation of 28 units. The sample data indicate an average of 107 units from 30 observations. Assuming a normal distribution, the z-test is appropriate because the population standard deviation is known. The z-statistic is computed as:
z = (x■ - µ) / (σ / √n) = (107 - 120) / (28 / √30) ≈ -13 / (28 / 5.477) ≈ -13 / 5.113 ≈ -2.543
The critical z-value for a two-tailed test at α = 0.05 is approximately ±1.96. Since -2.543 < -1.96, we reject the null hypothesis, indicating that the new employee’s average output is significantly different from the historical mean at the 5% level.
In the second scenario, the battery life is evaluated against the expected mean of 900 days with a known standard deviation of 150 days. With a sample of 40 batteries, the sample mean is 942 days. The z-test applies here as well. The test statistic is:
z = (x■ - µ) / (σ / √n) = (942 - 900) / (150 / √40) ≈ 42 / (150 / 6.3246) ≈ 42 / 23.693 ≈ 1.775
The critical z-value at a 5% significance level is ±1.96. Since 1.775 is less than 1.96, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the battery lifetime significantly differs from 900 days.
The third case examines whether boys in a municipal children’s home are underweight. The population mean weight is µ = 85 pounds, standard deviation σ = 11.6 pounds, and the sample includes 25 boys with a mean weight of 80.94 pounds. The z-test here is appropriate as population standard deviation is known:
z = (x■ - µ) / (σ / √n) = (80.94 - 85) / (11.6 / √25) ≈ -4.06 / (11.6 / 5) ≈ -4.06 / 2.32 ≈ -1.75
Since |z|= 1.75 < 1.96, we fail to reject the null hypothesis at the 0.05 significance level. Thus, the data do not provide sufficient evidence to support the claim that the boys are underfed.
The fourth scenario involves testing a principal’s claim that her students are of above-average intelligence.
The sample mean IQ is 107.5, with the population mean at 100 and population standard deviation of 15. The sample size is 24. The z-test is applicable:
z = (x■ - µ) / (σ / √n) = (107.5 - 100) / (15 / √24) ≈ 7.5 / (15 / 4.898) ≈ 7.5 / 3.058 ≈ 2.45
The critical z-value for a one-tailed test at α = 0.05 is approximately 1.645. Since 2.45 > 1.645, we reject the null hypothesis and conclude there is sufficient evidence to support the principal's claim that her students' mean IQ is above 100.
In the final scenario, the class's vocabulary test score is evaluated against the population mean of 68 with a standard deviation of 13. The sample mean is 65, with 19 students. The z-test is:
z = (x■ - µ) / (σ / √n) = (65 - 68) / (13 / √19) ≈ -3 / (13 / 4.359) ≈ -3 / 2.98 ≈ -1.01
As |z|= 1.01 < 1.96, we fail to reject the null hypothesis at the 0.05 significance level, suggesting the class’s score is not significantly different from the typical mean.
Overall, these hypothesis tests demonstrate how statistical analysis can inform judgments about real-world questions. When interpreting results, it is crucial to consider the context and whether the assumptions underlying each test are met. In cases where null hypotheses are rejected, it indicates the observed differences are unlikely to be due to random variation alone, pointing toward genuine effects or differences.
References
Casella, G., & Berger, R. L. (2002). Statistical Inference (2nd ed.). Thomson Learning.
Devore, J. L. (2011). Probability and Statistics for Engineering and the Sciences (8th ed.). Brooks/Cole.
Fisher, R. A. (1925). Statistical Methods for Research Workers. Oliver and Boyd.
Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W.H. Freeman.
Wasserman, L. (2004). All of Statistics: A Concise Course in Statistical Inference. Springer.
Agresti, A., & Finlay, B. (2008). Statistical Methods for the Social Sciences (4th ed.). Pearson.
Hogg, R. V., & Tanis, E. A. (2005). Probability and Statistical Inference (7th ed.). Pearson.
Rice, J. A. (2006). Mathematical Statistics and Data Analysis (3rd ed.). Brooks/Cole.
Gibbons, J. D., & Chakraborti, S. (2011). Nonparametric Statistical Inference (5th ed.). CRC Press.
Lehmann, E. L., & Romano, J. P. (2005). Testing Statistical Hypotheses (3rd ed.). Springer.