Question

Advertisers contract with Internet service providers and search engines to place ad on websites. They pay a fee based on the number of potential customers who click on their ad. Unfortunately, click fraud—the practice of someone clicking on and ad solely for the purpose of driving up advertising revenue--- has become a problem. Forty percent of advertisers claim they have been a victim of click fraud ( business Week, March 13, 2006). Suppose a simple random sample of 380 advertisers will be taken to learn more about how they are affected by this practice.A.What is the probability that the sample proportion will be within +/-.04 of the population experiencing click fraud?B.What is the probability that the sample reapportion will be greater that .45?

Answer 1

Answer:

b. P(ρ^>0.45) = 0.0233

Step-by-step explanation:

Hello!

To calculate the asked probabilities you need to first summarize the information given in the problem.

Study variable:

X: "Number of advertisers affected by click fraud"

n= 380 advertisers

ρ= 0.4 (probability of suffering click fraud)

ρ^ (sample proportion)

To calculate the probability of the sample proportion to take any number, you'll have to apply the Central Limit Theorem to approximate it's distribution to normal, that way

ρ^≈N(ρ;([ρ*(1-ρ)]/n)

Under this distribution, we can use de statistic Z= (ρ^-ρ)/√[ρ*(1-ρ)]/n ≈ N(0;1) to calculate the wanted probabilities.

a.

b.

P(ρ^>0.45) ⇒ 1 - P(ρ^≤0.45) ⇒ 1 - P(Z≤(0.45 - 0.4)/√[0.4*(1-0.4)]/380)

⇒ 1 - P(Z≤ 1.99) = 1 - 0.9767 = 0.0233

I hope you have a SUPER day!

Answer 2

Probability of an event is the measure of its chance of occurrence. The needed probabilities of the considered cases are:

• The probability that the sample proportion will be within +/-.04 of the population experiencing click fraud is 0.8836 = 88.36% approx.
• The probability that the sample proportion will be greater that .45 is 0.025 = 2.5% approx.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

$X \sim N(\mu, \sigma)$

(X is following normal distribution with mean $\mu$ and standard deviation $\sigma$)

then it can be converted to standard normal distribution as

$Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)$

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

$P(Z \leq z) = P(Z < z) )$

Also, know that if we look for Z = z in z tables, the p value we get is

$P(Z \leq z) = \rm p \: value$

For the given case, we can model the situation with normal distribution as:

$\hat{p}\sim N(p; \dfrac{p(1-p)}{n})\\\\\rm As\: p = 40\% = 0.4, and \: n = 380,\: thus,\\\\\hat{p}\sim N(\mu = 0.45; \sigma^2 = \dfrac{0.45 \times 0.55}{380}\approx 0.00065)$

where, $\hat{p}$ = sample proportion, and p = mean of sample proportion, which is equal to the population's proportion's average(given 40% = 0.4).

(it can be modeled by normal distribution as the sample size is large enough)

The needed probability for case A is for $\hat{p}$  to be within +/- 0.04 of the population experiencing click fraud.

That can be symbolically written as: $P(0.4-0.04 \leq \hat{p} \leq 0.4 + 0.04) = P(0.36 \leq \hat{p} \leq 0.44)$

Rewriting it gives us:

$P(0.36 \leq \hat{p} \leq 0.44) = P(\hat{p} \leq 0.44) - P(\hat{p} \leq 0.36)$

Using standard normal distribution conversion of the distribution of the sample proportion, we get the needed probability as:

$P(\hat{p} \leq 0.44) - P(\hat{p} \leq 0.36) \approx P(Z =\dfrac{\hat{p} - \mu}{\sigma} \leq \dfrac{0.44 - 0.4}{\sqrt{0.00065}} ) - P(Z \leq \dfrac{0.36-0.4}{\sqrt{0.00065}})\\\\P(\hat{p} \leq 0.44) - P(\hat{p} \leq 0.36) \approx P(Z \leq 1.57) - P(Z \leq -1.57)$

using the z-tables, the p-value for Z = 1.57 is 0.9418, and for Z = -1.57 is 0.0582, thus,

$P(\hat{p} \leq 0.44) - P(\hat{p} \leq 0.36) \approx P(Z \leq 1.57) - P(Z \leq -1.57)\\P(\hat{p} \leq 0.44) - P(\hat{p} \leq 0.36) \approx 0.9418 - 0.0582 = 0.8836$

Thus, $P(0.4-0.04 \leq \hat{p} \leq 0.4 + 0.04)) \approx 0.8836$

The probability that the sample proportion will be greater than 0.45 is written symbolically as:

$P(X > 0.45)$

It can be rewritten as: $P(X > 0.45) = 1 - P(X \leq 0.45)$

using the standard normal distribution, the considered probability is converted to:

$P(\hat{p} > 0.45) = 1 - P(\ahat{p} \leq 0.45) \approx 1 - P(Z = \dfrac{\hat{p} - \mu}{\sigma}} \leq \dfrac{0.45 - 0.4}{\sqrt{0.000065}})\\P(\hat{p} > 0.45) \approx 1 - P(Z \leq 1.96) =1-0.975 = 0.025$

Thus, the needed probabilities are evaluated to:

• The probability that the sample proportion will be within +/-.04 of the population experiencing click fraud is 0.8836 = 88.36% approx.
• The probability that the sample proportion will be greater that .45 is 0.025 = 2.5% approx.

Learn more about standard normal distribution here:

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