Respuesta :
Answer:
[tex]z=\frac{0.66 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=-2.230[/tex]
[tex]p_v =2*P(z-2.230)=0.0257[/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, and we can said that at 5% of significance the proportion of first-year students say that this goal is important is different from 0.73 at 5% of significance
Step-by-step explanation:
Data given and notation
n=200 represent the random sample taken
X=132 represent the number of first-year students say that this goal is important
[tex]\hat p=\frac{132}{200}=0.66[/tex] estimated proportion of first-year students say that this goal is important
[tex]p_o=0.73[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
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 differes from 0.73.:
Null hypothesis:[tex]p=0.73[/tex]
Alternative hypothesis:[tex]p \neq 0.73[/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{0.66 -0.73}{\sqrt{\frac{0.73(1-0.73)}{200}}}=-2.230[/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-2.230)=0.0257[/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, and we can said that at 5% of significance the proportion of first-year students say that this goal is important is different from 0.73 at 5% of significance
Answer:
Step-by-step explanation:
Hello!
The objective is to test id the proportion of first-year college students that identified "being very well-off financially" as an important personal goal is different from 73%, symbolically: p ≠ 0.73
The variable of interest is:
X: number of first-year students at a state university that considers "being very well-off financially" an important personal goal in a sample of 200.
The sample proportion is p'= 132/200= 0.66
The statistic hypotheses are:
H₀: p = 0.73
H₁: p ≠ 0.73
α: 0.05
This is a two-tailed test, using the approximation of the standard normal distribution, the critical values are:
[tex]Z_{\alpha /2}= Z_{0.025}= -1.965[/tex]
[tex]Z_{1-\alpha /2}= Z_{0.975}= 1.965[/tex]
Then you'll reject the null hypothesis if [tex]Z_{H_0}[/tex] ≤ -1.965 or if [tex]Z_{H_0}[/tex] ≥ 1.965
And you'll not reject the null hypothesis if -1.965 < [tex]Z_{H_0}[/tex] < 1.965
[tex]Z_{H_0}= \frac{p'-p}{\sqrt{\frac{p(1-p)}{n} } } = \frac{0.66-0.73}{\sqrt{\frac{0.73*0.27}{200} } } = -2.32[/tex]
Since the value of [tex]Z_{H_0}[/tex] is less than -1.965, the decision is to reject the null hypothesis.
Then at a 5% significance level, there is enough evidence to conclude that the proportion of first-year students at the state university that considers "being very well-off financially" an important personal goal is different from 73%
I hope this helps!
