Mathematics In the Modern World Module 4 Exercose 4 PDF

Preview

Summary

EXERCISE 31. D co. has an average gross sales of 37 million per week from their products in all outlets with a standard deviation of 6 million. An area manager found out that the average gross sales from the 32 outlets under his area is 34 million per week. Is the mean sales of the products in 32 br...

Description

EXERCISE 3

1. D.G co. has an average gross sales of 37 million per week from their products in all outlets with a standard deviation of 6 million. An area manager found out that the average gross sales from the 32 outlets under his area is 34.8 million per week. Is the mean sales of the products in 32 branches significantly different from the mean sales in all outlets? Use ∝ = 5% a. Perform the 5 steps in testing hypotheses with two-tailed or two- sided tests. b. Perform the 5 steps in testing hypotheses with one tailed or one - sided tests. And N= 45 branches. (All other given are the same) TABULAR VALUE OR CRITICAL Z-VALUE (C) Test type

Level of significance α=% (0.01)

α = 5%(0.05) α = 10%(0.1) α = 2.5%(0.0.25)

One-tailed or one-sided test

+ 2.33

+ 1.645

+ 1.28

+1.96

Two-tailed or two-sided test

+2.575

+1.96

+ 1.645

+2.33

(1A) SOLUTION: STEP 1: Ho = There is no significant difference between the mean sales of the 32 branches from the mean sales in all outlets. (

μ=37 ¿

Ha = There is a significant difference between the mean sales of the 32 branches from the mean sales in all outlets.

(μ ≠37)

STEP 2: TYPE OF TEST: Z test (n>30, population variance is known), two-tailed or non-directional test STEP 3: CRITICAL VALUE: with the use of table, ∝ = 5%, two-tailed test, the critical value of z = ±1.96 STEP 4: Given: X = 34.8

μ = 37 n = 32

σ =6

z=

´ ( X−μ ) √n

z=

σ

( 34.8−37) √ 32 6

z=

(−2.2 )5.657 6

z=

−12.46 6

z = ±2.07 STEP 5: Z-value does not fall within the accepted region of Ho Decision: Reject Ho Conclusion: Ha - There is a significant difference between the mean sales of the 32 branches from the mean sales in all outlets.

(1B) SOLUTION: STEP 1: Ho = There is no significant difference between the mean sales of the 45 branches from the mean sales in all outlets. (

μ=37 ¿

Ha = The mean sales from the 45 branches is lower than the mean sales in all outlets.

(μ ≠37) STEP 2: TYPE OF TEST: Z test (n>30) one-tailed STEP 3: CRITICAL VALUE: with the use of table, ∝ = 5%, one-tailed test, the critical value of z = ±1.645 STEP 4: Given: X = 34.8

μ = 37 n = 45

σ =6 z=

´ ( X−μ ) √n

z=

σ

(34.8−37 ) √ 45 6

z=

(−2.2 )6.708 6

z=

−14.76 6

z = ±2.46 STEP 5: Z-value is within the rejection region of Ho Decision: Reject Ho Conclusion: Ha - The mean sales from the 45 branches is lower than the mean sales in all outlets.

2. The subject coordinator wants to find out if the webcasting methods of teaching social sciences subjects are more effective than the existing standard method of teaching. Two classes of approximately equal intelligence were selected. From one class, the teacher selected students in random as an experimental group for webcasting methods, and from the other class, the teacher selected another sample of students for an existing standard method of teaching as a controlled group. After one trimester, the same tests were given to the students. The mean scores are shown in the following table. Method of Teaching

Mean Score

Standard Deviation

Class Size

E-learning and Web Casting

x = 82.3

12

n1 = 42

Existing Standard Method

x = 78.5

9

n2 = 38

Test the hypothesis that the web casting methods of teaching social sciences subjects is more effective than the existing standard method of teaching. Use α = 0.05.

(2) SOLUTION: STEP 1: Ho = There is no significant difference between the web casting methods of teaching social sciences subjects and the existing standard method of teaching

( x´ = x´ )

Ha = Web casting methods of teaching social sciences subjects is more effective than the

(x ≠ x´ )

existing standard method of teaching STEP 2:

TYPE OF TEST: Z test, n>30, one-tailed STEP 3: CRITICAL VALUE: with the use of table, α = 0.05. one-tailed test, the critical value of z = ±1.645

x 1−¿ x

√

STEP 4:

2

( n1−1) s + ( n 2−1 ) s 2 2 1

n1+ n2−2 ¿ 82.3 −78.5

√

2

2

√

( 42− 1 )12 +( 38−1 ) 9 1 1 + ) ( 42 38 42 + 38 −2

2

(

√

1 1 + ) n1 n2 3.8

√

( 4 1 ) 144+ ( 3 7 ) 81 ( √ 0.023 + 0.026 ) 78

3.8 √ 114.115(√ 0. 049)

3.8 3.8 10.682 ( 0.221) 10.682 (0.221 )

z= 1.61

STEP 5: Z-value of 1.61 is within the accepted region which is 1.645. Decision: Do not reject Ho Conclusion: Ho - There is no significant difference between the web casting methods of teaching social sciences subjects and the existing standard method of teaching

´ ( x´ =x)...