Question

Clark Heter is an Industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the night shirt than on the day shift. The mean number of units produced by a sample of 54 day-shift workers was 331. The mean number of units produced by a sample of 60 night-shift workers was 338. Assume the population standard deviation of the number of units produced on the day shift Is 19 and 26 on the night shift. At the 0.01 significance level, is the number of units produced on the night shift larger? a. Is this a one-talled or a two-talled test? One-talled test Two-talled test b. State the decision rule. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) The decision rule is to reject HO:pDay Night if z

Answer 1

Z-test for two samples

The z test applied when the two groups are compared. This test used when both the population variances values are known, sample sizes are larger than 30 and the distribution of both the populations follows normal. The observations in both the groups are not dependent on each other.

Let

Day = day shift mean

I Night = night shift mean

T Night = night shift sample mean = 338

Day = day shift sample mean = 331

n Night = night shift sample size = 60

n Day = day shift sample size = 54

a.

The statistical hypothesis is:

Ho : Jl Day <Night v/s Hp : Jl Day > Night (claim)

It is a One-tailed test.

b.

At α=0.01, for a one-tailed test, Z(crit) = 2.33

The decision rule is to reject $H_0$ ,if Z > Z(crit)

c.

Test Statistic,

IN - ID = ~ N(0,1)

338-331 紧+環

  

= 1.652127

*1.65

d.

Z(test) < Z(crit) so the test falls in the non-rejection region. Do not reject the null hypothesis $H_0$ .

There is not sufficient evidence to support the claim that the number of units produced on the night shift is larger than the number of units produced on the day shift.