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A social scientist believed that less than 30 percent of adults in the united states watch 15 or fewer hours of television per week. To test the belief, the scientist randomly selected 1,250 adults in the united states. The sample proportion of adults who watch 15 or fewer hours of television per week was 0. 28, and the resulting hypothesis test had a p-value of 0. 61. The computation of the p-value assumes which of the following is true?

Answer :

The computation of the p-value that is true is; The population proportion of adults who watch 15 or fewer hours of television per week is less than 0.30.

What is the p-value of the hypothesis?

Let the proportion of adults watching television less than or equal to 15% be  x

30 percent of adults in the united states watch 15 or fewer hours of television per week. Thus;

Null Hypothesis; H₀: x = 30% = 0.30

Alternative Hypothesis; H₁ : x < 30% , or x < 0.30

Formula for the z-score is:

z = p' - [√{p₀(1 - p₀)/n}]

where;

p' = 0.28, p₀ = 0.30, p₁ = 0.70

Using online p-value from z-score calculator we have;

p(z < -1.543) = 0.061

If we assume 10% level of significance, i.e p = 0.10

Thus, p-value 0.061 < 0.10 and we reject Null hypothesis and accept the alternative hypothesis.

Finally, we conclude that the proportion of adults watching television less than or equal to 15% <  30% or 0.30'

Read more about Confidence Intervals at; https://brainly.com/question/16362968

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