Question

In a recent year, 73% of first-year college students responding to a national survey identified "being very well-off financially" as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the α=0.05 significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value, 73%?

Answer 1

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:

`Z_(\alpha /2)= Z_(0.025)= -1.965`

`Z_(1-\alpha /2)= Z_(0.975)= 1.965`

Then you'll reject the null hypothesis if
`Z_(H_0)` ≤ -1.965 or if
`Z_(H_0)` ≥ 1.965

And you'll not reject the null hypothesis if -1.965 <
`Z_(H_0)` < 1.965

`Z_(H_0)= \frac{p'-p}{\sqrt{(p(1-p))/(n) } } = \frac{0.66-0.73}{\sqrt{(0.73*0.27)/(200) } } = -2.32`

Since the value of
`Z_(H_0)` 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!

Answer 2

`z=\frac{0.66 -0.73}{\sqrt{(0.73(1-0.73))/(200)}}=-2.230`

`p_v =2*P(z-2.230)=0.0257`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` 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

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

`\hat p=(132)/(200)=0.66` estimated proportion of first-year students say that this goal is important

`p_o=0.73` is the value that we want to test

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 true proportion differes from 0.73.:

Null hypothesis:
`p=0.73`

Alternative hypothesis:
`p \\eq 0.73`

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{(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{0.66 -0.73}{\sqrt{(0.73(1-0.73))/(200)}}=-2.230`

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=0.05`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z-2.230)=0.0257`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` 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