Question

You work for a lobby group that is trying to convince the government to pass a new law. Before embarking on this, your lobby group would like to know as much as possible about the level of community support for the new law.

Your colleague, based on his research into community opinion on related matters, proposes that 32% of the community support the law. You decide to survey 100 people, and find that 27% of this survey support the law.

a)Based on the assumption that the population proportion is 32%, calculate the z-score of the sample proportion in your survey. Give your answer as a decimal to 2 decimal places.

z =

b)Determine the proportion of the standard normal distribution that lies to the left of this z-score. That is, determine the area to the left of this z-score in the standard normal distribution. You may find this standard normal table useful. Give your answer as a percentage to 2 decimal places.

Area =  %

c)Denote by x% the percentage proportion you calculated in part b). Consider the following five potential conclusions:

A: There is a chance of x% that your friend is correct, that the true population proportion is 32%.

B: If your colleague is correct and the true population proportion is 32%, then x% of all samples will produce a sample proportion of 27% or lower.

C: If your colleague is correct and the true population proportion is 32%, then x% of all samples will produce a sample proportion of 27% or higher.

D: There is a chance of x% that the true population proportion is 32% or lower.

E: There is a chance of x% that the true population proportion is 32% or higher.

Select the statement that can be inferred from your findings:

A
B
C
D
E

Answer 1

a)

Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   27                  
Sample Size,   n =    100                  
                          
Sample Proportion ,    p̂ = x/n =    0.270                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0466                  
Z Test Statistic = ( p̂-p)/SE = (   0.2700   -   0.32   ) /   0.0466   =   -1.07

..................

b)

area to the left of this z-score = P(Z<-1.07)=0.1419     [excel function =NORMSDIST(z)]

so, area = 14.19%

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c)

C: If your colleague is correct and the true population proportion is 32%, then x% of all samples will produce a sample proportion of 27% or higher.