According to a CBS News survey 40 of Americans do not eat br

According to a CBS News survey, 40% of Americans do not eat breakfast. A sample of 30 high school students found 16 had skipped breakfast that day. Use the .01 significance level to check whether high school students are more likely to skip breakfast.

NEEDS TO BE IN EXCEL FORMAT

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Solution

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   <=   0.4
Ha:   p   >   0.4
As we see, the hypothesized po =   0.4      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.533333333      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.089442719      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    1.490711985      
          
As this is a right tailed test, then, getting the right tailed p value by typing

=1-NORMSDIST(1.490711985)

We get the p value of
          
p =    0.068018564  
  
Comparing p >0.01, we   FAIL TO REJECT THE NULL HYPOTHESIS.      

Thus, there is no significant evidence that high school students are more likely to skip breakfast. [CONCLUSION]