Question

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes.

A. Can we conclude that the assembly time using the new method is faster at the 1% significance level?

B. The new method will be implemented only if the assembly time is at least 90 seconds faster than the old method. Otherwise, the costs of the change are not worth the returns. Modify the null and alternate hypotheses of “part a” to account for this information? Complete the test and state your conclusion.

Answer 1

Here present method requires 42.3 minutes to assemple a cart.

Mean assembly time = 40.6 minutes

standard deviation of the sample = s = 2.7 minutes

sample size n = 24

A. Here

Null Hypothesis : H₀ : = 42.3 min

Alternative Hypothesis : H_a : < 42.3 min

standard error of sample mean = s/sqrt(n) = 2.7/sqrt(24) = 0.5511

Test statistic

t = (40.6 - 42.3)/0.5511 = -3.0845

Here for dF = 24 -1 = 23 and alpha = 0.01

t_critical = 2.8073

Here t > t_critical so we will reject the null hypothesis and can conclude that the new method is faster at the 1% significance level.

B. here

Null Hypothesis : H₀ : = 1.5 min

Alternative Hypothesis : H_a : > 1.5 min

Here difference in sample means =

the test statistic

t = [() - ()]/ se₀

= [(42.3 - 40.6) - 1.5] / 0.5511

= 0.363

Here critical value of test statistic = t_critical = 2.80

so here t < t_critical so we fail ro reject the null hypothesis and claim that there is not at least 90 seconds faster difference in the assembly time