Answer :
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 notation
n=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 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:[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 Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since 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 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 [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.
Final answer:
The student tests the claim that at least 55% of adults skip breakfast using a hypothesis test for a proportion, with a null hypothesis that the true proportion is at least 55%. A p-value lower than the significance level indicates the null hypothesis should be rejected suggesting sufficient evidence against the claim.
Explanation:
The question involves conducting a hypothesis test to check the claim made from a survey. A dietitian wants to test the claim that at least 55% of adults do not eat breakfast at least three days a week. The student's test at a 10% significance level (α = 0.10) revealed that 50% of the adults in the sample of 80 skipped breakfast three days a week or more. To assess this claim, we'll use a hypothesis test for a proportion.
To perform the test, we start by setting up the null (H0) and alternative (Ha) hypotheses:
H0: p ≥ 0.55 (The proportion of adults who skip breakfast at least three days a week is at least 55%.)
Ha: p < 0.55 (The proportion of adults who skip breakfast at least three days a week is less than 55%.)
Next, we calculate the test statistic using the sample proportion (50%) and the hypothesized proportion (55%). After finding the test statistic, we look up the corresponding p-value in a Z-distribution table (since the sample size is large enough to approximate the distribution of the sample proportion with a normal distribution). If the p-value is smaller than the significance level α, we reject the null hypothesis.
If the p-value was 0.0004 and we were testing at the 5% level (α = 0.05), as seen in the provided information, we would reject the null hypothesis because p-value < α. However, since our actual significance level for this question is α = 0.10, if the obtained p-value was the same (0.0004), we would still reject the null hypothesis. Therefore, there is sufficient evidence to conclude that fewer than 55% of adults skip breakfast at least three days a week.
