In a poll of 500 adults in July 2010 270 of those polled sai

In a poll of 500 adults in July 2010, 270 of those polled said that schools should ban sugary snacks and soft drinks. Complete parts a and b below. a. Do a majority of adults (more than 50% support a ban on sugary snacks and soft drinks? Perform a hypothesis test using a significance level of 0.05 State the null and alternative hypotheses. Note that p is defined as the population proportion of people who believe that schools should ban sugary foods. OA Ho: p 0.50 Ha p 4 0.50 OB. Ho: p 0.54 Ha p> 0.54 OC. Ho: p 0.50 OD. Ho: p 0.50 Ha p 0.50 Find the value of the test statistic z. z Round to two decimal places as needed.) Find the value of the corresponding p-value for this test statistic z p-value Round to three decimal places as needed.) Do you reject or not reject the null hypothesis? n Reiert H

Solution

a)

A majority is p > 0.50, which is the claim and the alternative hypothesis. Hence,as the null hypothesis is the opposite of the null,

Ho: p < 0.50
Ha: p > 0.50 [ANSWER]

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b)

Formulating the null and alternatuve hypotheses,          
          
Ho:   p   <=   0.5
Ha:   p   >   0.5

As we see, the hypothesized po =   0.5      
Getting the point estimate of p, p^,          
          
p^ = x / n =    0.54      
          
Getting the standard error of p^, sp,          
          
sp = sqrt[po (1 - po)/n] =    0.02236068      
          
Getting the z statistic,          
          
z = (p^ - po)/sp =    1.788854382   [ANSWER, TEST STATISTIC Z]

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As this is a    1   tailed test, then, getting the p value,  
          
p =    0.036819135   [answer, p value]

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As, significance level =    0.05      

Comparing p < significance value, we   REJECT THE NULL HYPOTHESIS.   [reject Ho]