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

`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%.

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=(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{(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{(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%.