Production engineers at a company believe that a modified layout on its assembly lines might increase average worker productivity (measured in the number of units produced per hour). However, before the engineers are ready to install the revised layout officially across the entire firm’s production lines, they would like to study the modified line’s effects on output. The following data represent the average hourly production output of 6 randomly sampled employees before and after the line was modified:

Employee 1 2 3 4 5 6

Before 43 45 47 46 48 44

After 42 46 48 50 46 51

At the .05 level of significance, can the production engineers conclude that the modified (after) layout has increased worker productivity? Conduct a complete and appropriate hypothesis test.

Let x = number of units produced per hour before the line was modified , y = number of units produced per hour after the line was modified.

The null and alternative hypotheses are :

H0: μbefore = μafter

H1: μbefore ≠ μafter

Step 1: . Calculate the difference (D = y − x) between the two observations on each pair.

See attachment for (D=y-x)

step 2: Calculate ∑D and ∑D2

From table we get ∑D = -10 & ∑D2 = 72

Step 3: Put all the value in test statistic "t"

t = D/√nD^2-D^2/n - 1

t =-10/√6.72-(-10)^2/6-1

t= -10√66.4

= -10/8.1486

= -1.227

Step 4: Compare tcal and ttab

At α = 0.05

t0.05 for 5 d.f. = 3.1634 (two tail test)

Hence | tcal | < t0.05

So, we accept the hypothesis.

So we conclude that the modified (after) layout has not increased worker productivity at 5% level of significance.

$53.50

step-by-step explanation:

you have to subtract the amount of money chuck earned ($72.50) by the cost of the gas for the mower ($11.25) and the new string for the trimmer he bought ($7.75).

72.50-11.25=61.25

and then subtract $61.25 by the second thing he bought (the new string for the trimmer)

61.25-7.75=53.5

so in total it's $53.50 (53 dollars and 50 cents)

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step-by-step explanation: