Population proportion *statistics*

[georgeh]

Homework Statement

Harley-Davidson motorcyles make up 14% of all the motorcyles registered in the United States. You plan to interview an SRS of 500 motorcyle owners.
a) what the approximate distribution of the proportion of your sample who owns harleys?
b) Is your sample likely to contain 20% or more who own Harleys? Is it likely to contain at least 15% Harley owners? Do normal probability calculations to answer these questions

Homework Equations

p-phat = count of success in the sample/n
p = mean of the sampling distribution
z =( p-hat - p)/ std.dev

The Attempt at a Solution

I believe p = .14. n = 500
I don't understand what is meant by "approximate distribution of the proportion of your sample who owns harleys".
If they mean p-hat, would p-hat =(500*14/100) / 500 ? bit that would give me .14..

 

[AlephZero]

Hint: The question says "Do normal probability calculations to answer these questions".

The "exact" probability distribution here is a Binomial distribution. How do you use a Normal distribution to approximate a Binomial distribution?

 

[Architeuthis Dux]

As a scientist, it is important to clarify any uncertainties or misunderstandings in the given content. In this case, it is not clear what is meant by "approximate distribution," so it would be helpful to ask for clarification from the source of the information. Without this information, it is difficult to provide a complete and accurate response.

Assuming that the statement is referring to the distribution of the proportion of Harley owners in the sample, we can use the given information to calculate the standard deviation of the sampling distribution. This can be done using the formula:

σ = √(p(1-p)/n)

Where p is the population proportion (0.14) and n is the sample size (500). This gives us a standard deviation of approximately 0.012 or 1.2%.

a) With this information, we can say that the approximate distribution of the proportion of Harley owners in the sample is a normal distribution with a mean of 0.14 and a standard deviation of 0.012.

b) To determine the likelihood of the sample containing a certain proportion of Harley owners, we can use the z-score formula:

z = (x - μ)/σ

Where x is the proportion we are interested in and μ is the mean of the sampling distribution (0.14).

For 20% Harley owners, the z-score would be (0.20 - 0.14)/0.012 = 5, which is extremely unlikely in a normal distribution. Similarly, for at least 15% Harley owners, the z-score would be (0.15 - 0.14)/0.012 = 0.83, which is also unlikely but more plausible than 20%.

In conclusion, based on the given information, it is unlikely that the sample will contain 20% or more Harley owners, but it is possible that it will contain at least 15% Harley owners. However, it is important to note that this is only an estimate and the actual proportion of Harley owners in the sample may vary.