## 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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2022-02-11T09:08:51+00:00
2022-02-11T09:08:51+00:00 1 Answer
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## Answers ( )

Answer:[tex]z=\frac{0.5 -0.55}{\sqrt{\frac{0.55(1-0.55)}{80}}}=-0.899[/tex]

[tex]p_v =P(z<-0.899)=0.184[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] 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 notationn=80 represent the random sample taken

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

[tex]p_o=0.55[/tex] is the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)

Concepts and formulas to useWe 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:[tex]p\geq 0.55[/tex]

Alternative hypothesis:[tex]p < 0.55[/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 Testis used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].Calculate the statisticSince we have all the info requires we can replace in formula (1) like this:

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

Statistical decisionIt’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=0.1[/tex]. The next step would be calculate the p value for this test.

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

[tex]p_v =P(z<-0.899)=0.184[/tex]

So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v>\alpha[/tex] 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.