In? 1997, a survey of 900900 households showed that 157157 of them use? e-mail.
ID: 3369319 • Letter: I
Question
In? 1997, a survey of
900900
households showed that
157157
of them use? e-mail. Use those sample results to test the claim that more than? 15% of households use? e-mail. Use a 0.05 significance level. Use this information to answer the following questions.
a. Which of the following is the hypothesis test to be? conducted?
A.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p less than 0.15H1: p<0.15
B.
Upper H 0 : p not equals 0.15H0: p?0.15
Upper H 1 : p equals 0.15H1: p=0.15
C.
Upper H 0 : p greater than 0.15H0: p>0.15
Upper H 1 : p equals 0.15H1: p=0.15
D.
Upper H 0 : p less than 0.15H0: p<0.15
Upper H 1 : p equals 0.15H1: p=0.15
E.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p greater than 0.15H1: p>0.15
F.
Upper H 0 : p equals 0.15H0: p=0.15
Upper H 1 : p not equals 0.15H1: p?0.15
b. What is the test? statistic?
zequals=nothing
?(Round to two decimal places as? needed.)
c. What is the? P-value?
?P-valueequals=nothing
?(Round to three decimal places as? needed.)
d. What is the? conclusion?
There
isis
sufficient evidence to support the claim that more than? 15% of households use? e-mail.There
is notis not
sufficient evidence to support the claim that more than? 15% of households use? e-mail.
e. Is the conclusion valid? today? Why or why? not?
A.
?Yes, the conclusion is valid today because the requirements to perform the test are satisfied.
B.
?No, the conclusion is not valid today because the population characteristics of the use of? e-mail are changing rapidly.
C.
You can make no decisions about the validity of the conclusion today.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.15
Alternative hypothesis: P > 0.15
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).
S.D = sqrt[ P * ( 1 - P ) / n ]
S.D = 0.01190
z = (p - P) / S.D
z = 2.05
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
Since we have a one-tailed test, the P-value is the probability that the z-score is more than 2.05.
Thus, the P-value = 0.02.
Interpret results. Since the P-value (0.02) is less than the significance level (0.05), we cannot accept the null hypothesis.
There is sufficient evidence to support the claim that more than? 15% of households use? e-mail.
B) ?No, the conclusion is not valid today because the population characteristics of the use of? e-mail are changing rapidly.