a. Assuming that the VP selects exactly 50 calculators to use in the demonstration, and using the Poisson distribution as an approximation of the binomial, what is the chance of getting at least three calculators that malfunction?

b. No calculators malfunctioning?

Solution

P (x ≥ 3) = 1 −¿ P (x ≤ 2)

= 1 −¿ [P (x =0) + P (x =1) + P (x =2)]

= 1 −¿ [0.13533 + 0.27066 +0.27066]

= 0.32335

1. Hypothesis Testing (Single Population)

Problem

General Electric has developed a new bulb whose design specifications call for a light output of 960 lumens compared to an earlier model that produced only 750 lumens. The company’s data indicate that the standard deviation of light output for this type of bulb is 18.4 lumens. From a sample of 20 new bulbs, the testing committee found an average light output of 954 lumens per bulb. At a 0.05 significance level, can General Electric conclude that its new bulb is producing the specified 960 lumen output?

Solution

σ = 18.4n = 20x = 954µ = 960α = 0.05µ = 960

H 0 µ = 960

H 1 µ < 960

Zc = - 1.65

Z test = x−µσ /√ n

= 954−96018.4/√ 20

= - 1.45

Answer

Do not rejectHO. The new bulb is meeting specifications.

2. Hypothesis Testing (Single Population)

Problem

BSNL provides telephone services in Coimbatore. According to the company’s records the average length of calls placed through the company is 11.44 minutes. The company wants to check if the mean length of the current calls is different from 11.44 minutes. A sample of 150 such calls placed through this company gave a mean length of 12.71 minutes with a standard deviation of 2.65 minutes. Can you conclude that the mean length of all current calls is different from 11.44 minutes? Use α = 0.05.

Solution

s = 2.65n = 150x = 12.71µ = 11.44α = 0.05

H 0 µ = 11.44

H 1 µ ≠ 11.44

Zc = ± 1.65

Z test = x−µσ /√ n

= 12.71−11.442.65/√ 150

= 5.87

Answer

Reject H 0. It is concluded that the mean length of current calls is different from 11.44 minutes.

3. Hypothesis Testing ( Two

who were trained using the new method and found average daily sales to be $ 7

Answer

a.) The chance of getting at least three calculators that malfunction is 32.33%b.) The chance of no calculators malfunctioning is 13.53%

06 and the sample standard deviation was $24.84. At alpha = 0.05, can the company conclude that average daily sales have increased under the new plan?

Solution

n1 = 16 n2 = 11 n = n1 + n2 = 27

x1 = 688 x2 = 206

s 1 = 32.63 s 2 = 24.84

α = 0.05

H 0 µ1 - µ2 = 0

H 1 µ1 - µ2 ≠ 0

T c = - 1.708

sp2 = s12(n1−1)+s22(n2−1)

n1+n2−2

= 32.632(15)+24.842(10)

25

= 885.64

T test= ( x1−x 2 )−(µ1−µ2)

√sp2∗√( 1n1

+ 1n2

)

= (688−706 )−(0)

√885.64∗√ ( 116

+ 111

)

4. populations)

Problem

A credit-insurance organization has developed a new high-tech method of training new sales personnel. The company sampled 16 employees who were trained the original way and found average daily sales to be $688 and the sample standard deviation was $32.63. They also sampled 11 employees