Question

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Questions 34 – 35: Residential Development A new residential development is being considered in a suburban district of a city. The question which the city planning committee would like to be answered is how big this residential development should be. They want to commission a survey of 400 people living in that district to know what they think. The district is quite cosmopolitan and it is assumed that the opinions on such issues are generally shared by adult people of similar age and gender does not play a part. It is well known that in that district there are 30% young professionals, 45% mid-age couples with children and 25% seniors. The market research company came up with a plan to have a sample of total size 400 made up of 120 young professionals, 180 mid-age couples ad 100 seniors chosen as a collection of 3 simple random samples

34. What is this sampling design called? a) Simple Random Sample. b) Cluster Sample. c) Stratified Random Sample. d) Systematic Sample. e) 3-Stage Sample.

11 35. It was found that 280 of the 400 people sampled, responded that they were in favour of a large development. What is the probability that 280 or less number of people would be in favour of a large development assuming that the developer contends that ¾ th of the people are in favour of this large development? Is the assertion of ¾ th of the people being in favour justified?

a) 0.0105; the assertion is not justified.

b) 0.0412; the assertion is justified.

c) 0.0570; the assertion is justified.

d) 0.5000; the assertion is not justified.

e) None of the above

Answer 1

34

The sampling design is called Stratified Random Sample

Stratified sampling

The researcher divides the population into separate groups, called strata. Then, a probability sample (often a simple random sample) is drawn from each group.

35)

Option A

Proportion of people in favour (p1) = 280/400 = 0.7

Count (n) = 400

Null and Alternate Hypothesis

H₀: p = ¾ = 0.75

H_a: p < 0.75

Test Statistic

t = (p1 – p₀) / {p₀*(1-p0)/n}^1/2 = -2.31

p-value = TDIST(2.31,400-1,1) = 0.0107

Result

Since the p-value is less than 0.05, we reject the null hypothesis.