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 33% of the community support the law. You decide to survey 100 people, and find that 39% of this survey support the law.

a)Based on the assumption that the population proportion is 33%, 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 right of this z-score. That is, determine the area to the right 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 33%.

B: If your colleague is correct and the true population proportion is 33%, then x% of all samples will produce a sample proportion of 39% or lower.

C: If your colleague is correct and the true population proportion is 33%, then x% of all samples will produce a sample proportion of 39% or higher.

D: There is a chance of x% that the true population proportion is 33% or lower.

E: There is a chance of x% that the true population proportion is 33% or higher.

Select the statement that can be inferred from your findings:

A
B
C
D
E

Answer 1

P- population proportion of the community support the law =33%=0.33

n = sample sizeof people decided survey =100

p^ =Sample proportion of people who support low =39%=0.39.

a)Based on the assumption that the population proportion is 33%, calculate the z-score of the sample proportion in your survey. Give your answer as a decimal to 2 decimal places.

So here z -score =1.28.

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b)Determine the proportion of the standard normal distribution that lies to the right of this z-score. That is, determine the area to the right 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.

Here use standard normal table woth z =1.28

So split z=1.28 as 1.2 and 0.08 .

Search 1.8 value in z- column and 0.08 value in z-row .Then write corresponding value in table .

That will be peobability P(Z < 1.28) Since normal distribution always given us less than (Left side ) probability .

Notw : Total area under curve = 1

           Left side area +Right side area = 1

         So right side area = 1- Left side area ==>>>

             P(Z > 1.28) = 1 - P( Z < 1.28).

Lets use table to find P( Z <1.28) .

So , Area at right side =P( Z>1.28) = 1- P(Z <1.28) .

and from table we get ,P( Z <1.28) = 0.8997

So Area at right side = P(Z > 1.28 ) = 1 - 0.8997 = 0.1003

Area = 10.03 %

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

We choose option C: If your colleague is correct and the true population proportion is 33%, then x% of all samples will produce a sample proportion of 39% or higher.