Question

According to a study by the American Pet Food Dealers association, 63 percent of U.S. households own pets. A report is being prepared for an editorial in the San Francisco Chronicle. As a part of the editorial a random sample of 300 households showed 210 own pets. Does this data disagree with the Pet Food Dealers Association data? Use a .05 level of significance.

Answer 1

`z=\frac{0.7 -0.63}{\sqrt{(0.63(1-0.63))/(300)}}=2.511`

`p_v =2*P(z>2.511)=0.012`

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 households with pets is different from 0.63. This data disagree deom the Pet food association

Explanation:

Data given and notation

n=300 represent the random sample taken

X=210 represent the households with pets

`\hat p=(210)/(300)=0.7` estimated proportion of households with pets

`p_o=0.63` 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 true proportion is 0.63:

Null hypothesis:
`p=0.63`

Alternative hypothesis:
`p \\eq 0.63`

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{0.7 -0.63}{\sqrt{(0.63(1-0.63))/(300)}}=2.511`

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>2.511)=0.012`

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 households with pets is different from 0.63. This data disagree deom the Pet food association