Determine the point estimate of the population proportion, the margin of error

Mathematics

1 answer:

ahrayia [7]3 years ago

8 0

Answer:

$\hat p=0.645\\\\ME=0.233\\\\x=645 \ individuals$

Step-by-step explanation:

-Given the boundaries as 0.412 and 0.878

- $\hat p$ is the point estimate for the population proportion and is calculated as follows:

$\hat p=\frac{Upper \ Bound+Lower \ Bound}{2}\\\\=\frac{0.878+0.412}{2}\\\\=0.645\\\\$

#The margin of error, ME can be calculated for the confidence intervals using the formula:

$ME=\frac{Upper \ Bound-Lower \ Bound}{2}\\\\=\frac{0.878-0.412}{2}\\\\=0.233$

#The number of individuals in the sample is the product of the point estimate and population size:

$\hat p=\frac{x}{n}\\\\x=\hat pn\\\\=0.645\times 1000\\\\=645$

Hence, there are 645 individuals in the sample.

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let X = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, p = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, n = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable X follows a Binomial distribution with parameter n = 55 and p = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$