Question

Does second-hand smoke increase the risk of a low birthweight? A baby is considered having low birth weight if he/she weighs less than 5.5 pounds at birth. According to the National Center of Health Statistics, about 7.8% of all babies born in the U.S. are categorized as low birthweight.
Suspecting that the national percentage is higher than 7.8%, researchers randomly select 1200 babies whose mothers had extensive exposure to second-hand smoke during pregnancy and find that 10.4% of the sampled babies are categorized as low birth weight.
Let p be the proportion of all babies in the U.S. that are categorized as "low birth weight." Give the null and alternative hypotheses for this research question.
H 0 : p = 0.078
H 0 : p = 0.078
H a : p > 0.078
H 0 : p = 0.104

Answer 1

Answer:

Null hypothesis:$p \leq 0.078$

The null hypothesis can be on this way: $p=0.078$, but is better put the complement of the alternative hypothesis.  

Alternative hypothesis:$p > 0.078$  

Step-by-step explanation:

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: "Suspecting that the national percentage is higher than 7.8%". So we want to check if the population proportion of babies with low weigth is higher than 0.078 or 7.8%, 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:$p \leq 0.078$

The null hypothesis can be on this way: $p=0.078$, but is better put the complement of the alternative hypothesis.    

Alternative hypothesis:$p > 0.078$  

And if we want to test this hypothesis we can use a z test for a proportion.

$\hat p =0.104$ represent the proportion on the sample.

n= 1200 represent the sample size

$p_o =0.078$ represent the value that we want to test.

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.104-0.078}{\sqrt{\frac{0.078(1-0.078)}{1200}}}=3.36$  

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 next step would be calculate the p value for this test.  

Since is a one side right tailed test the p value would be:  

$p_v =P(z>3.36)=0.00039$

So the p value obtained was a very low value and using a significance level for example $\alpha=0.05$ we have $p_v$ 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 low weight babies is significantly higher than 0.078 or 7.8%.