Lab A12 - StatsCh9

1 Lab Activity A12 : Stats Ch 9 Paid days off and the single-sample t test: Part 1. The number of paid days off (e.g., vacation, sick leave) taken by eight employees at a small local business is compared to the national average. You are hired as a consultant by the new business owner to help her determine how many paid days off she should provide. In general, she wants to set some standard for her employees and for herself, and so she first wants to determine whether the paid time off taken by her employees is different from the national average. Your research on paid days off reveals that the national average is 15 days. The data for the eight local employees during the last fiscal year are: 10, 11, 8, 14, 13, 12, 12, and 27 days. The mean is 13.375 and the standard deviation is 5.805. 1. Write alternative and null hypothesis statements for your research, in statistical notation. HINT: think about whether you should do a one-tailed test or a two-tailed test. H0: u=15 H1: u  15 One tailed 2. Calculate the appropriate t statistic. -0.793 3. What is the degrees of freedom for this sample, and the t -critical(s) if alpha = .05? Df=7 t-critical= 1.895 4. What is your conclusion: should you reject the null or fail to reject the null? What does this conclusion mean in plain English? Fail to reject the null hypothesis. This means that our t statistic was less than the t critical value, resulting in the failing to reject null hypothesis. 5. Look up the p -value for this t -statistic using this online calculator: https://www.socscistatistics.com/pvalues/tdistribution.aspx Report the p -value, rounded to two decimals (remember that if your p -value is less than . 001, you can write this as p < .001). Is this p -value less than alpha? P value= 0.23 The p value is less than alpha. Part 2. After further investigation, you discover that one of the data points, 27 days, was actually the owner’s number of paid days off. You re-do your analyses after removing this data point and obtain a new mean of 11.429 and a new standard deviation of 1.988. Given your new mean, you now hypothesize that this sample has a lower number of paid days off compared to the national average. Conduct a new hypothesis test, adjusting for this new information. 6. What are the new hypothesis statements, in statistical notation? H0: u>15

2 H1: u<15 7. What is the new t -statistic? -4.76 8. What is/are the critical value(s) if alpha = .05? 1.943 9. What is the statistical decision now? What does this conclusion mean in plain English? We reject the null hypothesis. This means that there is a definite relationship between the average hours compared to the national average. Also, our t-stat was greater than our critical value. 10. Look up the p -value for your new t -statistic using this online calculator: https://www.socscistatistics.com/pvalues/tdistribution.aspx Report the p -value, rounded to two decimals (remember that if your p -value is less than . 001, you can write this as p < .001). Is this p -value less than alpha? P<.001 The p value is less than alpha Part 3. Errors and Effect Sizes. 11. Calculate Cohen’s d and describe it as small, medium, or large. D= 0.45 Based on the chart below, I would describe it as medium 12. In part 1, what kind of error did we commit (Type I or Type II error)? Explain your answer. We committed a type 2 error because we failed to reject the null when it was actually false based on part 2. FORMULAS Formula for t -statistic: Formula for Cohen’s d for a single-sample t -test: Cohen ' sd = | M − μ s | Cohen’s d Effect Size Overlap 0.2 Small 85% t statistic = ( M − μ ) s √ N