Question

1. It turns out the true proportion of people who play with the ACME Spinner is 0.23. For his statistics project at ACME University, Johnny surveyed 100 people and asked them if they ever played with the ACME Spinner. Of these, 26 responded “yes”.
   1. If the true proportion really is 0.23 (or 23%), what is the exact chance or probability of getting a sample proportion of 0.26 or more in a sample of size 100? Use the binomial distribution.
   2. If the true proportion really is 0.23 (or 23%), use the central limit theorem to approximate the chance or probability of getting a sample proportion of 0.26 or more in a sample of 100. Compare with your answer to part e.
   3. If the true proportion really is .23, what is the exact chance of observing a sample proportion of .26 or more in a sample of size 200? Again, use the binomial distribution. Compare your answer with that of part e.
   4. If the true proportion really is .23, use the central limit theorem to approximate the chance of observing a sample proportion of .26 or more in a sample of size 200. Compare your answer with that of parts f and g. Explain the differences.

Answer 1

a)

The following information is provided:

The population proportion of success is p = 0.23, also, 1 - p = 1 - 0.23 = 0.77,

and the sample size is n= 100.

We need to compute Pr(X 26)

Pr(X 26) = 1 - Pr[ X   25]

This implies that

now Pr(X 26) = 1 - Pr[ X   25] = 1 - 0.7283 = 0.2717

b)

The following information about the Binomial distribution is provided:

The population proportion of success is p = 0.23, and the sample size is n= 100.

We need to compute Pr(X≥26):

The population mean is computed as:

and the population standard deviation is computed as:

Therefore, we get that

so then, we conclude that

Pr(X≥26)≈0.2762,

c)

The following information is provided:

The population proportion of success is p = 0.23, also, 1 - p = 1 - 0.23 = 0.77,

and the sample size is n= 200.

We need to compute Pr(X 26)

Pr(X 26) = 1 - Pr[ X   25]

Therefore, we get that

This implies that

Pr(X≤25)=0.0001.

now

Pr(X 26) = 1 - Pr[ X   25] = 1 - 0.0001 = 0.9999

D)

The population proportion of success is p = 0.23,

and the sample size is n= 200.

We need to compute Pr(X≥26):

The population mean is computed as:

and the population standard deviation is computed as:

Therefore, we get that

so then, we conclude that Pr(X≥26)≈0.9997,

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