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stepladder [879]
3 years ago
6

Production engineers at a company believe that a modified layout on its assembly lines might increase average worker productivit

y (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.
Mathematics
1 answer:
marshall27 [118]3 years ago
7 0

Answer:

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.

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Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting
slega [8]

Answer:

(a) <em>λ</em> = 2.

(b) P (X = 0) = 0.1353; P (X = 1) = 0.2706;

    P (X = 2) = 0.2706; P (X = 3) = 0.1804

(c) P (Delay Problems) = 0.1431.

Step-by-step explanation:

Let <em>X</em> = number of arrivals at the drive-up teller window.

The average number of arrivals at the drive-up teller window per minute is,

<em>p</em> = 0.4 customers/ minute.

(1)

Compute the expected number of customers at the drive-up teller window in <em>n</em> = 5 minutes as follows:

E(X)=\lambda\\=np\\=5\times 0.4\\=2

Thus, the mean number of customers that will arrive in a five-minute period is <em>λ</em> = 2.

(2)

The random variable <em>X</em> follows a Poisson distribution with parameter λ = 2.

The probability mass function of <em>X</em> is:

P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...

Compute the probability of exactly 0 arrivals in 5 minutes as follows:

P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353

Compute the probability of exactly 1 arrivals in 5 minutes as follows:

P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706

Compute the probability of exactly 2 arrivals in 5 minutes as follows:

P(X=2)=\frac{e^{-2}2^{2}}{2!}=\frac{0.1353\times 4}{2}=0.2706

Compute the probability of exactly 3 arrivals in 5 minutes as follows:

P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804

Thus, the values are:

P (X = 0) = 0.1353

P (X = 1) = 0.2706

P (X = 2) = 0.2706

P (X = 3) = 0.1804

(3)

Delays occur in the service time if there are more than three customers arrive during any five-minute period.

Compute the probability that there are more than 3 customers as follows:

P (X > 3) = 1 - P (X ≤ 3)

              =1-\sum\limits^{3}_{x=0}{\frac{e^{-2}2^{x}}{x!}}\\=1-(0.1353+0.2706+0.2706+0.1804)\\=1-0.8569\\=0.1431

Thus, the probability that delays will occur is 0.1431.

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<em><u>Solution:</u></em>

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