Question

The Heldrich Center for Workforce Department found that 45% of Internet users received more than 10 email messages per day. Recently, a similar study on the use of email was reported. The purpose of the study was to see whether the use of email increased.

a. Formulate the null and alterative hypotheses to determine whether an increase occurred in the proportion of Internet users receiving more than 10 email messages per day.

b. If a sample of 420 Internet users found 208 receiving more than 10 email messages per day, what is the p-value?

c. Using α =.05, what is your conclusion?

d. Apply the critical value method when a = .01.

Answer 1

a) Null hypothesis:
`p \leq 0.45`

Alternative hypothesis:
`p > 0.45`

b)
`p_v =P(z>1.854)=0.0319`

c) 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 internet users recieving more than 10 email messages per day is not significantly higher than 0.45

d) For this case since the new value for the significance is
`\alpha=0.1` and we are conducting a right tailed test we need to look in the normal tandard distribution for a quantile that acucmulates 0.99 of the area on the left and 0.01 on the right and this critical value is:

`z_(cric)= 2.326`

And since our calculated value is lower than 2.326 we FAIL to reject the null hypothesis at 1% of significance.

Explanation:

Data given and notation

n=420 represent the random sample taken

X=208 represent the internet users recieving more than 10 email messages per day

`\hat p=(208)/(420)=0.495` estimated proportion of internet users recieving more than 10 email messages per day

`p_o=0.45` 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)

Part a Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that we have an increase os the proportion of interest, so the system of hypothesis are:

Null hypothesis:
`p \leq 0.45`

Alternative hypothesis:
`p > 0.45`

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`.

Part b

Calculate the statistic

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

`z=\frac{0.495 -0.45}{\sqrt{(0.45(1-0.45))/(420)}}=1.854`

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>1.854)=0.0319`

Part c

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 internet users recieving more than 10 email messages per day is not significantly higher than 0.45

Part d

For this case since the new value for the significance is
`\alpha=0.1` and we are conducting a right tailed test we need to look in the normal tandard distribution for a quantile that acucmulates 0.99 of the area on the left and 0.01 on the right and this critical value is:

`z_(cric)= 2.326`

And since our calculated value is lower than 2.326 we FAIL to reject the null hypothesis at 1% of significance.