Question

The national percent of adults who have a laptop computer is 62%. In San Francisco, a simple random sample of 100 adults is taken and 69% of these adults have a laptop. Is this strong evidence that more adults in SF have laptops than the national average?

a) State the null hypothesis and the alternative in terms of a box model.
b) Find z and p values.
c) What do you conclude?

Answer 1

a) Null hypothesis:
`p\leq 0.62`

Alternative hypothesis:
`p > 0.62`

b)
`z=\frac{0.69 -0.62}{\sqrt{(0.62(1-0.62))/(100)}}=1.442`

`p_v =P(Z>1.442)=0.235`

c) The p value obtained was a high low value and using the significance level assumed for example
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have a laptop is not significantly higher than 0.62 .

Explanation:

1) Data given and notation

n=100 represent the random sample taken

X represent the adults that have a laptop

`\hat p=0.69` estimated proportion of adults that have a laptop

`p_o=0.62` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim the proportion is higher than 0.62 the national proportion:

Null hypothesis:
`p\leq 0.62`

Alternative hypothesis:
`p > 0.62`

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`.

3) Calculate the statistic

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

`z=\frac{0.69 -0.62}{\sqrt{(0.62(1-0.62))/(100)}}=1.442`

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

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

`p_v =P(Z>1.442)=0.235`

The p value obtained was a high low value and using the significance level assumed for example
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who have a laptop is not significantly higher than 0.62 .