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 200 incoming email messages at this college, the sample proportion of these messages that are spam would exceed .75. 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 200 messages contains 75% 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 200 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.

$P(p>0.75)=P\left ( \frac{p-0.8}{\sqrt{\frac{0.8(1-0.8)}{200}}}> \frac{0.75-0.8}{\sqrt{\frac{0.8(1-0.8)}{200}}}\right ) =1-\Phi (-1.77)=\Phi (1.77)=0.9616\\$

2.

3.

$P(p>0.75)=P\left ( \frac{p-0.75}{\sqrt{\frac{0.75(1-0.75)}{200}}}> \frac{0.75-0.75}{\sqrt{\frac{0.75(1-0.75)}{200}}}\right ) =1-\Phi (0)=0.5\\$

Yes. Since Probability of proportion of the messages that are spam exceed 0.75 in this situation is reduced.

4.

$P(p>0.75)=P\left ( \frac{p-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}}> \frac{0.75-0.80}{\sqrt{\frac{0.80(1-0.80)}{100}}}\right ) =1-\Phi (-1.25)\\ =\Phi (1.25)=0.8944\\ Hence~the~probability~is~reduced.\\$

$5.~P(p>0.75)=P\left ( \frac{p-0.77}{\sqrt{\frac{0.77(1-0.77)}{200}}}> \frac{0.75-0.77}{\sqrt{\frac{0.77(1-0.77)}{200}}}\right ) =1-\Phi (-0.6721)\\ =\Phi (0.6721)=0.7492\\ Hence~the~probability~is~reduced.\\$