A popular blog reports that 60% of college students log onto facebook on a daily

Question

A popular blog reports that 60% of college students log onto facebook on a daily basis. The Dean of students at a certain university thinks that the proportion may be different at her university. She polls a simple random sample of 205 students, and 134 of them report they log into facebook daily. Can you conclude that the proportion of students who login to facebook daily differs 0.60? Use the .05 level of sifnificance. Find the P-Value and State a Conclusion:

Select the right answer(s).

A) There is enough evidence to conclude that the proportion of students who log into facebook daily differs from 0.60

B) P-value = .9416

C) P- Value =.0584

D) P- Value = .1168

E) There is not enough evidence to conclude that the proportion of students who login to facebook daily differs from .60

F) P-value = .0094

Explanation / Answer

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   =   0.6
Ha:   p   =/=   0.6
As we see, the hypothesized po =   0.6      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.653658537      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.034215957      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    1.568231358      
          
As this is a    2   tailed test, then, getting the p value,  
          
p =    0.116827149 = 0.1168   [ANSWER, P VALUE]  

As P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.      

Hence, the P value and conclusion are

OPTION D: D) P- Value = .1168 [ANSWER]

OPTION E: E) There is not enough evidence to conclude that the proportion of students who login to facebook daily differs from .60 [ANSWER]

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