4. A multinational food company conducted a consumer survey in different countri

Question

4. A multinational food company conducted a consumer survey in different countries to assess the level of demand for a new food product. From a random sample of 1000 consumers in Africa, 25% indicated that they would purchase the new food product. In Asia, a random sample of 800 consumers found that only 20% would purchase the new food product.

a) Required: Using an appropriate test at the 5% level of significance, determine whether there is a significant difference between the proportions of consumers who would purchase the new food product in the two continents. In your answer to this question, you are required to:

1. State the null and alternative hypotheses.

2. Determine the critical value of z for this test.

3. Calculate the test statistic.

4. Draw an appropriate conclusion.

b) The weekly demand for a product is normally distributed with a mean of 2000 and a standard deviation of 200. Calculate the probability that the demand for the product on a given week is:

1. Less than 1750

2. More than 2450

3. Between 2000 and 2450

Explanation / Answer

Q.4 (a) 5% level of significance,

1. State the null and alternative hypotheses.

Null : H0 : Proportion of consumers purchase new food product are same in both continent. p1 = p2

ALternative : Ha : Proportion of consumers purchase new food product are different in both continent. p1 p2

2. Pooled estimate p = (1000 * 0.25 + 800 * 0.20) / (1000 + 800) = 0.2278

Test statistic

Z = (p1 -p2 )/ sqrt [ p * (1-p) * (1/n1 + 1/n2 )] = (0.25 - 0.20)/ sqrt [ 0.2278 * 0.7722 * (1/1000 + 1/800)]

Z = 0.05/ 0.01989 = 2.51

3. Critical value of Z = 1.96

4. We shall reject the null hypothesis and can conclude that there is difference between Proportion of consumers purchase new food product.

(b) Weekely demand mean = 2000

Staandard deviation = 200

1. Less than 1750 = Pr (weekly demand < 1750; 2000 ; 200)

Z = (1750 - 2000)/200 = -1.25

Pr (weekly demand < 1750; 2000 ; 200) = (-1.25) = 0.1056

where is the normal cumulative distribution function.

2. More than 2450.

Pr (weekly demand > 2450; 2000 ; 200) = 1 - Pr (weekly demand < 2450; 2000 ; 200)

Z = (2450 - 2000)/ 200 = 2.25

Pr (weekly demand > 2450; 2000 ; 200) = 1 - (2.25) = 1 - 0.9878 = 0.0122

3. Between 2000 and 2450

Pr (2000 < weekly demand < 2450; 2000 ; 200) = Pr (weekly demand < 2450; 2000 ; 200) - Pr(weekly demand < 2450; 2000 ; 200)

Z = (2450 - 2000)/ 200 = 2.25

Pr (2000 < weekly demand < 2450; 2000 ; 200) = Pr (weekly demand < 2450; 2000 ; 200) - Pr(weekly demand < 2000 2000 ; 200) = 0.9878 0.5 = 0.4878

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