Respuesta :
Answer:
a) [tex]z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649[/tex]
b) For this case we need to find a critical value that accumulates [tex]\alpha/2[/tex] of the area on each tail, we know that [tex]\alpha=0.01[/tex], so then [tex] \alpha/2 =0.005[/tex], using the normal standard table or excel we see that:
[tex] z_{crit}= \pm 2.58[/tex]
Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.
Step-by-step explanation:
Data given and notation
n=420+174=594 represent the random sample taken
X=174 represent the number of yellow peas
[tex]\hat p=\frac{174}{594}=0.293[/tex] estimated proportion of yellow peas
[tex]p_o=0.23[/tex] is the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level
Confidence=99% or 0.99
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 true proportion of yellow peas is 0.23:
Null hypothesis:[tex]p=0.23[/tex]
Alternative hypothesis:[tex]p \neq 0.23[/tex]
When we conduct a proportion test we need to use the z statisitc, 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{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649[/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=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z>3.649)=0.00026[/tex]
So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.
b) Critical value
For this case we need to find a critical value that accumulates [tex]\alpha/2[/tex] of the area on each tail, we know that [tex]\alpha=0.01[/tex], so then [tex] \alpha/2 =0.005[/tex], using the normal standard table or excel we see that:
[tex] z_{crit}= \pm 2.58[/tex]
Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.
