Question

Q One. A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They believe that the firm’s professionals volunteer an average of more than 15 hours per month. If this is not the case, they will institute an incentive program to increase community involvement. A random sample of 24 professionals reported the following number of hours: 12 13 14 14 15 15 15 16 16 16 16 16 17 17 17 18 18 18 18 19 19 19 20 21 1. Write the null and alternative hypotheses. 2. In this context, describe the Type I error possible. How might such an error impact the software development firm? 3. In this context, describe the Type II error possible. How might such an error impact the software development firm? 4. What is the value of the test statistic? 5. What is the associated P-value? 6. State your conclusion using α = .05. 7. For a more accurate determination, top management wants to estimate the average number of hours volunteered per month by their professional staff to within one hour with 99% confidence. How many randomly selected professional employees would they need to sample? 8. Suppose 40 professional employees are randomly selected. This sample yields a mean of 15.2 hours and a standard deviation of 1.8 hours. Find a 95% confidence interval and interpret.

Answer 1

Soln

1)

Null and Alternate Hypothesis

H₀: µ = 15 (Mean Number of Hours = 15)

H_a: µ > 15 (Mean Number of Hours > 15)

2)

If we wrongly reject the true null hypothesis, then it would be a Type 1 Error. (ie we reject the null hypothesis when in fact it is true)

3)

If we fail to reject the false null hypothesis, then it would be a Type 2 Error.

4)

Sample Mean = 16.625

Std Dev = 2.22

n = 24

Std Error = 2.22/24^1/2 = 0.454

Test Statistic

t = (16.625 – 15)/0.454 = 3.58

5)

P-value = TDIST(t,n-1,1) = TDIST(3.58,23,1) = 0.00079

6)

Alpha = 0.05

Since the p-value is less than 0.05, we reject the null hypothesis.

7)

Margin of Error = 1

Alpha = 0.01

ZCritical = 2.58

Margin of Error = Z_Critical ­* S/n^1/2

n = (Z_Critical ­* S/Margin of Error)² = (2.58*2.22)² = 32.88 ~ 33

8)

Alpha = 0.05

ZCritical = 1.96

95% CI = Mean +/- ZCritical * Std Dev/n^1/2 = 15.2 +/- 1.96*1.8/40^1/2 = {14.64,15.76}

Interpretation

We are 95% confident that the true mean lies in the interval {14.64,15.76}