Question

Spam: A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that the percentage of emails that are spam is greater than 69% ? Use both =α0.01 and =α0.05 levels of significance and the P-value method with the TI-84 Plus calculator.

Answer 1

a) No

B) Yes

Explanation:

Calculating the p-value, we have;

z = (p-bar) -p/√(p(1-p)/n)

But p-bar = 365/500

= 0.73

Therefore,

z = 0.73 -0.69/√0.69(1-0.69)/500

= 0.04/√0.2139/500

= 0.04/√0.0004278

= 0.04/0.02068

= 1.93

p-value = p(z ≥1.93) = 0.0268

(a) Can you conclude that the percentage of emails that are spam is greater than 69% ?

No, because the p- value is greater than α. That is p-value⊃ 0.01

(b) Can you conclude that the percentage of emails that are spam is greater than 69%

Yes, since p-value ∠ 0.05

Answer 2

Complete Question:

Spam: A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a=0.05 levels of significance and the -value method with the table.

(a) State the appropriate null and alternate hypotheses.

(b) Compute the -value.

(c) At the a=0.01, can you conclude that greater than 69% of emails are spam?

(d) At the a=0.05, can you conclude that greater than 69% of emails are spam?

Answer and step-by-step explanation:

a)

The required hypothesis are

H₀:
`\mu` = 0.69

H₁:
`\mu` > 0.69

additional solutions are attached in the image below