Question

According to CNN, 55% of all u.S. smartphone users have used their gPS capability to get directions (CNN.com, May 2016). Suppose a provider of wireless telephone service in Canada wants to know if gPS usage by its customers differs from u.S. smartphone users. The company collects usage records for this year for a random sample of 547,192 of its customers and determines that 302,050 of these customers have used their telephone’s gPS capability this year. use this information to test the Canadian company’s claim at a 5 .01. Discuss the practical significance of the results.

Answer 1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H₀ : p = 0.55

H_a : p $\neq$ 0.55

n = 547192

x = 302050

$\hat p$ = x / n = 302050 / 5471952 = 0.552

P₀ = 0.55

1 - P₀ = 1 - 0.55 = 0.45

z = $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.552 - 0.55 / [$\sqrt$(0.55 * 0.45) / 547192]

= 2.974

Test statistic = 2.974

P(z > 2.974) = 1 - P(z < 2.974) = 1 - 0.9985 = 0.0015

P-value = 2 * 0.0015 = 0.0030

$\alpha$ = 0.01

P-value < $\alpha$

Reject the null hypothesis .

There is enough evidence to suggest that the GPS usage by its customers differs from u.S. smartphone users .