Question

One major factor in low productivity is the amount of time wasted by workers. Wasted time could include things such as time spent waiting for more material and equipment, cleaning up mistakes, and performing any other activity not related to production. In a project designed to examine this problem, a consultant took a survey of 200 workers in companies that were classified as successful (on the basis of their latest annual profits) and another 200 workers from unsuccessful companies. The amount of time (in hours) wasted during a standard 40-hour workweek was recorded for each worker. Do these data provide enough evidence at the 5% significance level to infer that the amount of time wasted in unsuccessful firms exceeds that of successful ones

Answer 1

Answer:

Part a: There exists enough statistical evidence at 1% level of significance to show that the amount of time wasted in unsuccessful firms exceeds that of successful ones.

Part b: 95% Confidence interval for how much more time is wasted in unsuccessful firms than in successful ones is (2.3104, 3.2496) hours.

Explanation:

Degrees of Freedom

df1 = n1 - 1 , df2 = n2 - 1 , df = n1 + n2 - 2

df = 398

Pooled Variance

The pooled variance is given as

$S_p^2=\frac{df_1SS_1+df_2SS_2}{df_1+df_2}$

By putting values in the formula,

Sp2 = 5.7401

Mean Squared Error Sm1-m2

$S_{m1-m2}=\sqrt{\frac{S_p^2}{n_1}+\frac{S_p^2}{n_2}}$

Putting values in the equation gives

Sm1-m2 = 0.2396

t-statistic

$t=\frac{(M_1-M_2)(\mu_1-\mu_2)}{S_{m1-m2}}$

By putting in the values,

t-statistic = t = -11.6034

p-value

The p value for the t-static -11.6034 and df=398, from the t tables

p is found as 0.

Decision

0 < 0.01

that is p-value is less than alpha.

Hence we Reject the null hypothesis.

Conclusion

There exists enough statistical evidence at 1% level of significance to show that the amount of time wasted in unsuccessful firms exceeds that of successful ones.

Part b:

We find 95% confidence interval for μ2 - μ1

From the given data, for 95% Confidence interval

α = 0.05,      α/2 = 0.025

From z tables for 95% confidence interval, z is given as 1.96.

Confidence interval is given by

$\bar{x_2}-\bar{x_1}\pm z\sqrt{\frac{s_2^2}{n_2}+\frac{s_1^2}{n_1}}\\7.8-5.02\pm 1.96\sqrt{\frac{3.09^2}{200}+\frac{1.39^2}{200}}\\= (2.3104, 3.2496)$

95% Confidence interval for how much more time is wasted in unsuccessful firms than in successful ones is (2.3104, 3.2496) hours.