Spam: A researcher reported that 71.8% of all email sent in a recent month was spam. A system manager at a large corporation bel
ieves 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.
2 answers:
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: = 0.69
H₁: > 0.69
additional solutions are attached in the image below
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Answer:
The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.
Step-by-step explanation:
79% of the fruit sold at Fred's Fruit stand has been any good. Test if there has been an increase.
At the null hypothesis, we test if there has been no increase, that is, the proportion is still of 79%, so:
At the alternative hypothesis, we test if there has been an increase, that is, more than 79% being good, so:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
0.79 is tested at the null hypothesis:
This means that
In a random sample if 475 pieces, 85 were bad.
So 475 - 85 = 390 were good, and:
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is the probability of finding a sample proportion above 0.8211, which is 1 subtracted by the p-value of Z = 1.66.
Looking at the z-table, z = 1.66 has a p-value of 0.9515.
1 - 0.9515 = 0.0485
The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.