Question

Nationally, 23% of Americans watch the nightly news at least once a week. In a random sample of 100 Portlanders, I find that 16% of the sample watched the nightly news once a week. Is this evidence that Portland’s news watching is different from the rest of the country?

9. Set up a hypothesis to test, and describe your hypothesis both in plain language, and in our H0 / HA notation.

10. Test the hypothesis you generated (maybe with a confidence interval).

11. Write a conclusion, both in our formal language of statistical hypothesis, and in plain language.

Answer 1

9)

H0: p= 0.23

Ha: p 0.23

10)

Margin of error = Z/2 * sqrt [ ( 1 - ) / n ]

= 1.96 * sqrt [ 0.16 * ( 1 - 0.16) / 100 ]

= 0.072

95% confidence interval = - E < p < +E

0.16 - 0.072 < p < 0.16 + 0.072

0.088 < p < 0.232

95% CI is ( 0.088 , 0.232 )

11)

Since Claimed proportion 0.23 contained in confidence interval, we fail to reject H0.

We conclude that we do not have sufficient evidnece to support the claim that Portland’s news

watching is different from the rest of the country.