Question

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 50.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 60 carts, using the new method, was 48.6 minutes, and the standard deviation of the sample was 2.8 minutes.

Using the 0.10 level of significance, can we conclude that the assembly time using the new method is faster?

a. What is the decision rule? (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Reject H₀: μ ≥ 50.3 and accept H₁: μ < 50.3 when the test statistic is (less than  equal to or  greater than) _______  

b. What is the value of the test statistic? (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

value of the test statistic:

c. What is your decision regarding H₀?

(Reject or Fail) to reject  H₀ .

Answer 1

Given that, sample size (n) = 60, sample mean = 48.6 minutes

and sample standard deviation (s) = 2.8 minutes

The null and alternative hypotheses are,

H0 : μ ≥ 50.3

H1 : μ < 50.3

This hypothesis test is a left-tailed test.

a) Degrees of freedom = 60 - 1 = 59

Using Excel find the t-critical value at 0.10 level of significance with 59 degrees of freedom as follows :

Excel Command : =TINV(2 * 0.10, 59) = 1.296

=> Critical value = -1.296 (negative since it is left-tailed test).

Decision Rule : Reject H0: μ ≥ 50.3 and accept H1: μ < 50.3 when the test statistic is less than or equal to -1.296.

b) Test statistic is,

=> Test statistic = -4.703

c) Since, test statistic = -4.703 < -1.296, we Reject H0.