Question

An estimated 78.1% of U.S. households had high-speed internet in 2013. A researcher wants to test this claim in his area because he believes that the estimate should be much higher. In a random sample of 510 households it was found that 408 had high-speed internet connections. Is there evidence to show that the percentage of U.S households in this researchers area with high- speed internet connection is greater than 78.1%. Let a = .05

What is the test statistic for this hypothesis test?

what are the critical values?

What is the p-value?

What is the decision for this hypothesis test?

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is

H₀ : p = 0.781

H_a : p > 0.781

= x / n = 408/510 = 0.800

P₀ = 0.781

1 - P₀ = 1-0.781 = 0.219

Test statistic = z

= - P₀ / [P_{0 *} (1 - P₀ ) / n]

=0.800-0.781 / [0.781*(0.219) /510 ]

= 1.038

P(z >1.038 ) = 1 - P(z < 1.038 ) = 0.1497

P-value = 0.1497

= 0.05   

  The significance level is α=0.05

The critical value for a right-tailed test is z_c​=1.64.

p = 0.1497 ≥ 0.05, it is concluded that the null hypothesis is not rejected.

There is not enough evidence to claim that the population proportion pp is greater than p_0​, at the α=0.05 significance level.