Question

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.

Answer 1

Answer:

$z=2.53$  

$p_v =P(z>2.53)=0.0057$  

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ 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 taken

X=540 represent the adults that said that schools should ban sugary snacks and soft drinks

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of adults that said that schools should ban sugary snacks and soft drinks

$p_o=0.5$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ 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:$p\leq 0.5$  

Alternative hypothesis:$p > 0.5$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$  

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 $\alpha=0.05$. 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:  

$p_v =P(z>2.53)=0.0057$  

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v<\alpha$ 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%.