You should not use Minitab to answer this question. Do the calculations by hand\

Question

You should not use Minitab to answer this question. Do the calculations by hand' (using a calculator) and show your working. The Science Council has a category of membership for Registered Science Technician (RSci Tech). Applicants for membership are assessed against a competency framework that includes 'B4: Take responsibility for completing tasks and procedures as well as using judgement within defined parameters'. One employer has noticed that, using the company's own assessment scale for this competency, historically their employees score 6.7. The employer wants to check the company assessment score for a group of their employees who already have RSciTech status. The data for the employees who have RSciTech status are summarised as follows: n = 16, sample mean x^bar = 8.6, sample standard deviation s = 2.31. What are the number of degrees of freedom and the critical value for a t-test applied to these data to check the company assessment score for the employees with RSciTech status? Hence calculate a 95% confidence interval for the population mean assessment score for the 'taking responsibility' competency for employees with RSciTech status.

Explanation / Answer

(A)
Number of degrees of freedom = 16 - 1 = 15
This is two tailed test. Considering the significance level of 0.05
critical value of t for df = 15 is +/-2.1315

(B)
Standard error SE = s/sqrt(n) = 2.31/sqrt(16) = 0.5775

Margin of error ME = t*SE = 2.1315*0.5775 = 1.2309

confidence interval = mean +/- ME = 8.6 +/- 1.2309 = (7.3691, 9.8309)

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