Question

Suppose 80% of the incoming email messages for a college’s computer system are spam.

1. Use the CLT to approximate the probability that in a random sample of 160 incoming email messages at this college, the sample proportion of these messages that are spam would exceed .70 .

2. Display your answer to question 1 as a shaded area in a well-labeled sketch.

3. After implementing a new spam blocker, if it turns out that a random sample of 160 messages contains 70% spam, would this constitute fairly convincing evidence that the (population) percentage of spam has been reduced from 80%? Explain, based on your calculation in question 1.

4. Would your answer to question 1 be larger, smaller, or the same if the sample size were 100 messages rather than 160 messages? Explain briefly.

5. Would your answer to question 1 be larger, smaller, or the same if the (population) proportion of spam messages were .77 rather than .80? Explain briefly.

Answer 1

1. The probability that in a random sample of 160 incoming email messages at this college, the sample proportion of these messages that are spam would exceed .70 .

PtX >0 70) _ PGrep;翯) ance B , )-P.rar»- Since E(p)=P,Va r p 0.70-p 0.70-0.80 (0.80×0.20)/160 =P(Z >-3.162) =1-P(Z > 3.162) =1-[1-P(Z <3.162)] = 0.9992

2.

Distribution Plot Normal, Mean=0.8, StDev=0.031622 14-T 0.9992 12 10 4 0.7

3)

X-P <070 =) Since E(p) = Pe P(X <070)= P Since E(p)=P,Va? (p ) = 0.70- P PQIn 0.70-0.80 (0.80×0.20)/160) P(Z <-3.162) =1-P(Z <3.162) = 1-0.9992=0.0008

The probabiltiy is very low So there is evidence that the population percentage of spam is reduced from 80%.

4. If the sample size n = 100 then

P(X > 070) = P X- P 0.70-FP O > Since E(p)=P,Var (p)= 0.70-P NPQ/100 0.70-0.80 10.80×0.20)/100 = P(Z >-2.5) =1-P(Z >2.5) =1-[1-P(Z <2.5)] = 0.9938

The probabilities of Question (1) and Question (4) are approximately same if the sample size were 100 messages rather than 160 messages

5.

X- P 0.77- FP P(X > 077) = P Since E(p)=P, Var (p)= 0.77-P PO/100 0.77-0.80 o.80xo.20)/160 = P(Z > 9487) = 1-P(Z > .9487) = 1-[1-P(Z < .9487)] = 0.82861

The answer to question 1 be smaller if the (population) proportion of spam messages were .77 rather than .80