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1997, a survey of 940 house holds showed that 15 of them use e-mail. Use those sample results to test the claim that more than 15 % of households use email. use a 0.05 significance level.
which of the following is the hypothesis test to be conducted

a. h0 :p=0.15
h1:p> 0.15

b h0: p<0.15
h1; p=0.15

c. ho>0.15
h1 =0.15

d. h0: p/= 0.15
h1 p=0.15

e h0 p=0.15
h1 p<0.15

f. H0:p =0.15
h1:p?=0.15

what is the test statistic? z=
what is the p-value
d. what is the conclusion?
a. There is not sufficient evidence to support the claim that more than 15% of households use e-mail.
b there is sufficient evidence to support the claim that more than 15%of households use email

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 email are changing rapidly.
c. you can make no decisions about the validity of the conclusion today.

  • statistics -

    Use a) for your hypotheses.

    You can try a proportional one-sample z-test since this problem is using proportions.

    Using a formula for a proportional one-sample z-test with your data included, we have:
    z = .16 - .15 -->test value (150/940) is .16) minus population value (.15) divided by
    √[(.15)(.85)/940] --> .85 represents 1-.15 and 940 is sample size.
    Finish the calculation.
    Use a z-table to find the p-value. The p-value is the actual level of the test statistic.

    Check a z-table for the critical value at .05 level of significance for a one-tailed test. Compare the test statistic you calculated to the critical value from the table. If the test statistic exceeds the critical value, reject the null and conclude p>0.15 (there is sufficient evidence to support the claim); if the test statistic does not exceed the critical value from the table, do not reject the null (there is not sufficient evidence to support the claim).

    I'll let you take it from here.

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