Question

A publisher reports that 49 % of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 200 found that 42 % of the readers owned a personal computer. Is there sufficient evidence at the 0.10 level to support the executive's claim? tep 1 of 7: State the null and alternative hypotheses Step 2 of 7: Find the value of the test statistic. Round your answer to two decimal places. Step 3 of 7: Specify if the test is one-tailed or two-tailed. ep 4 of 7: Determine the P-value of the test statistic. Round your answer to four decimal places. Step 5 of 7: Identify the value of the level of significance 6 of 7: Make the decision to reject or fail to reject the null hypothesis. Step 7 of 7: State the conclusion of the hypothesis test.

Answer 1

1) Null hypothesis:
`p=0.49`

Alternative hypothesis:
`p \\eq 0.49`

2)
`z=\frac{0.42 -0.49}{\sqrt{(0.49(1-0.49))/(200)}}=-1.98`

3) Bass on the alternative hypothesis we are conducting a bilateral test.

4)
`p_v =2*P(Z<-1.98)=0.0477`

5) For this case
`\alpha =0.1` is given.

6) Since we are conducting a bilateral test we have two critical values. On this case we need to find on the normal standard distribution two values that accumulates 0.05 of the area on each tail. And for this case the values are:

`z_(crit)= \pm 1.64`

So we reject the null hypothesis when
`z_(calc)>1.64 , z_(calc)<-1.64`

And if our calculated value is between (-1.64, 1.64) we FAIL to reject the null hypothesis.

7) If we compare the p value and the significance level given
`\alpha=0.1` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 10% of significance the proportion of the readers owned a personal computer is different from 0.49.

Explanation:

Data given and notation

n=200 represent the random sample taken

`\hat p=0.42` estimated proportion of the readers owned a personal computer

`p_o=0.49` is the value that we want to test

`\alpha=0.1` represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

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

1) Null and alternative hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion is different from 0.49:

Null hypothesis:
`p=0.49`

Alternative hypothesis:
`p \\eq 0.49`

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`.

2) Calculate the statistic

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

`z=\frac{0.42 -0.49}{\sqrt{(0.49(1-0.49))/(200)}}=-1.98`

3) Specify if the test is one-tailed or two-tailed

Bass on the alternative hypothesis we are conducting a bilateral test.

4) P value

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=0.1`. 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<-1.98)=0.0477`

5) Identify the value of the level of significance

For this case
`\alpha =0.1` is given.

6) Make the decision to reject or fail to reject the null hypothesis

Since we are conducting a bilateral test we have two critical values. On this case we need to find on the normal standard distribution two values that accumulates 0.05 of the area on each tail. And for this case the values are:

`z_(crit)= \pm 1.64`

So we reject the null hypothesis when
`z_(calc)>1.64 , z_(calc)<-1.64`

And if our calculated value is between (-1.64, 1.64) we FAIL to reject the null hypothesis.

7) State the conclusion of the hypothesis test.

If we compare the p value and the significance level given
`\alpha=0.1` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that at 10% of significance the proportion of the readers owned a personal computer is different from 0.49.