Question

A dietitian read in a survey that at least 55% of adults do not eat breakfast at least three days a week. To verify this, she selected a random sample of 80 adults and asked them how many days a week they skipped breakfast. A total of 50% responded that they skipped breakfast at least three days a week. At α=0.10 , test the claim.

Answer 1

`z=\frac{0.5 -0.55}{\sqrt{(0.55(1-0.55))/(80)}}=-0.899`

`p_v =P(z<-0.899)=0.184`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.1` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of adults who do not eat breakfast at least three days a week is NOT significantly less than 0.55.

Explanation:

Data given and notation

n=80 represent the random sample taken

`\hat p=0.5` estimated proportion of adults who do not eat breakfast at least three days a week

`p_o=0.55` is the value that we want to test

`\alpha=0.1` represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is at least 0.55, so the system of hypothesis would be:

Null hypothesis:
`p\geq 0.55`

Alternative hypothesis:
`p < 0.55`

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`.

Calculate the statistic

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

`z=\frac{0.5 -0.55}{\sqrt{(0.55(1-0.55))/(80)}}=-0.899`

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=0.1`. The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

`p_v =P(z<-0.899)=0.184`

So the p value obtained was a very high value and using the significance level given
`\alpha=0.1` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 10% of significance the proportion of adults who do not eat breakfast at least three days a week is NOT significantly less than 0.55.