Question

In: Statistics and Probability

1. The Manager of YoursTruly Agency decides to see whether a series of professional development seminars...

1. The Manager of YoursTruly Agency decides to see whether a series of professional development seminars improves on work days per year. Some randomly selected employees take the seminars. Another sample of randomly selected employees without training are part of the control group. One year after the seminars, the figures shown in the accompanying table are available:

	Seminar Group Days Worked	No Seminar Group Days Worked
Mean	213	203
Standard Deviation	19	23
Sample Size (n)	16	25

State your null and alternative hypotheses:

a. Using data from the question above, calculate your t-score – which involves first calculating the standard of error for the difference, s.e. D, then dividing the difference in means by this value (see formulas in your text; assume independent, unequal variance.)

1.b. Estimate the p-value for your t-score (using the simplified formula for degrees of freedom k equal to n₁+n₂-2, where n are the sample sizes.) You may do this using the Excel function=tdist(t, d.f., 1 tail)

1c. 1c. Is the difference of means statistically different from zero?

Is it practically different from zero?

In other words, what can you tell the manager about this experiment? Should s/he institute the new training procedure? Does the sample size matter in your advice (would it be different if the samples were 50 each)?

Expert Solution

The null and alternative hypothesis are as follows.:

The mean working days of the seminar group employees is same as the mean working days of the no seminar group employees.

The mean working days of the seminar group employees is greater the mean working days of the no seminar group employees.

a.

The formula for t-score assuming independent samples and unequal variance is

Now substitute the values in the formula.

The value of t-score is 1.512.

1.b.

The p-value for the t-score is given as

The p-value is 0.07 using the excel function = tdist(1.512, 39,1).

1.c.

The p-value is greater than 0.05. It means that the difference between means is not statistically different from zero. It defines that the new training procedure has no effect but as the p-value is closer to 5% and so it is advisable to check with other samples with different sample sizes. Sample size matter in this advice as the results will come more precise and reliable with higher sample sizes.

orchestra answered 2 years ago

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