Complete question :
Online photos: A poll surveyed 1765 internet users and found that 865 of them had posted a photo or video online. Can you conclude that less than half of internet users have posted photos or videos online? Use the α 0.01 level of significance and the P-value method with the TI-84 calculator.
1. State the appropriate null and alternate hypotheses.
2. Compute the P-value.
Answer:
H0 : p = 0.5
H1 : p < 0.5
P value = 0.200454
Step-by-step explanation:
Given that :
Claim : Less than half internet users have posted photos or videos online.
Sample Proportion (p) = 865 / 1765 = 0.4900
The null hypothesis : H0 : p = 0.5 (that is half of internet users have posted photos online.
The alternative hypothesis : H1 : p < 0.5 (that is less than half of internet users have posted photo or video ;
P = 0.5
q = 1 - 0.5 = 0.5
The test statistic :
Z = p - p0 / sqrt(p0 * (q/n))
Z = (0.49 - 0.5) / sqrt(0.5 * (0.5/1765))
Z = - 0.01 / sqrt(0.5 *0.0002832)
Z = - 0.01 / sqrt(0.0001416)
Z = - 0.01 / 0.0119013
Z = - 0.8402443
The corresponding p value from the z statistic score at α = 0.01
= 0.200454
