In a poll of 1000 adults in July 2010, 540 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.
Accepted Solution
A:
Answer:[tex]z=2.53[/tex] [tex]p_v =P(z>2.53)=0.0057[/tex] The p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.Step-by-step explanation:1) Data given and notation n=1000 represent the random sample takenX=540 represent the adults that said that schools should ban sugary snacks and soft drinks[tex]\hat p=\frac{540}{1000}=0.54[/tex] estimated proportion of adults that said that schools should ban sugary snacks and soft drinks[tex]p_o=0.5[/tex] is the value that we want to test[tex]\alpha=0.05[/tex] represent the significance levelConfidence=95% or 0.95z would represent the statistic (variable of interest)[tex]p_v[/tex] represent the p value (variable of interest) 2) Concepts and formulas to use We need to conduct a hypothesis in order to test the claim that majority of adults (more than 50%) support a ban on sugary snacks and soft drinks, the system of hypothesis are: Null hypothesis:[tex]p\leq 0.5[/tex] Alternative hypothesis:[tex]p > 0.5[/tex] When we conduct a proportion test we need to use the z statistic, and the is given by: [tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1) The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].3) Calculate the statistic Since we have all the info requires we can replace in formula (1) like this: [tex]z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53[/tex] 4) Statistical decision It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis. The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test. Since is a one side right tailed test the p value would be: [tex]p_v =P(z>2.53)=0.0057[/tex] So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.