Question

A recent FCC report stated that80% of all U.S. adults are internet users. In a poll run recently by PEW research of 3.011 adults, 2,478 said that they use the internet. A) Test the claim of the FCC at the .05 level. B) If the FCC claim were that more than 80% of all U.S. adults use the internet, how would that change your alternative hypothesis and your reject region?

Answer 1

A)

p(hat)= 2478/3011=0.82298

n= 3011

........two tail test

alpha= 0.05..........rejection region z score below -1.96 and above 1.96

Test Statistics

Z=(p(hat)-P)/sqrt(p*q/n)

=(0.82298-0.8)/sqrt(0.8*0.2/3011)

=3.1527

P-value=1-P(Z<3.1527)= 0.000808677 (this p-value is for one tail to get p-value for two tail we need to multiply it by 2.)

rounded p-value=2*0.000807=0.00162

Decision making

P-value ? Alpha Reject H0

0.00162<0.05

Alternate hypothesis become true so null hypothesis become false

FCC report stated that 80% of all U.S. adults are internet users is wrong.

there is enough evidence to support the claim that adults uses internet is different from 80%.

B)

p(hat)= 2478/3011=0.82298

n= 3011

H0: P= 0.8

Ha: P> 0.8 ..........right tail test

alpha= 0.05................rejection region is z score above 1.645

Test Statistics

Z=(p(hat)-P)/sqrt(p*q/n)

=(0.82298-0.8)/sqrt(0.8*0.2/3011)

=3.1527

P-value=1-P(Z<3.1527)= 0.00080

rounded p-value=0.0008

Decision making

P-value ? Alpha Reject H0

0.0008<0.05

there is enough evidence to support the claim that adults uses internet is more than 80 %.

.......................

if you have any doubt ask in comment give thumbs up if you like work