## 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 ran

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.

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3 months 2022-02-11T09:08:51+00:00 1 Answer 0 views 0

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

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

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.

Step-by-step 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:

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

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

Calculate the statistic

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

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

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.