Respuesta :
Answer:
a) H0: p = .53 vs. Ha: p < 0.53
b) [tex]p_v =P(Z>-2.429)=0.0075[/tex]
c) The p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the true proportion is less than 0.53.
Step-by-step explanation:
Part a
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
On this case the claim that they want to test is: "If the true proportion of American families own stocks or stock funds is less than 0.53 (The value ten years ago)". So we want to check if the population proportion actually is less than 0.53, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.
Null hypothesis:[tex]p \geq 0.53[/tex]
The null hypothesis can be on this way: [tex]p=0.53[/tex], but is better put the complement of the alternative hypothesis.
Alternative hypothesis:[tex]p < 0.53[/tex]
And the correct option would be:
d. H0: p = 0.53 vs. Ha: p < 0.53
Part b. b. Assume the Investment Company Institute sampled 300 American families to estimate that the percent owning stocks or stock funds was 46% in 2012. What is the p-value for your hypothesis test?
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the proportion is lss than 0.53:
Null hypothesis:[tex]p\geq 0.53[/tex]
Alternative hypothesis:[tex]p < 0.53[/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{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429[/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 assumed is [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a left taild test the p value would be:
[tex]p_v =P(Z<-2.429)=0.0075[/tex]
Part c. At α=.01, what is your conclusion?
The p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/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 1% of significance the true proportion is less than 0.53.