Question

Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. It is nearly impossible for most people to reliably list all the people they know, but using a mathematical model, social analysts estimate that, on average, a US adult has social ties with 634 people. A survey of 1700 randomly selected US adults who are Internet users finds that the average number of social ties for the Internet users in the sample was 669 with a standard deviation of 732. Does the sample provide evidence that the average number of social ties for an Internet user is significantly different from 634, the hypothesized number for all US adults (α=0.10)? Show all details of the test.

Answer 1

`t=(669-634)/((732)/(√(1700)))=1.971`

`p_v =2*P(t_((1699))>1.971)=0.049`

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Explanation:

Data given and notation

`\bar X=669` represent the sample mean

`s=732` represent the sample standard deviation

`n=1700` sample size

`\mu_o =68` represent the value that we want to test

`\alpha=0.` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:

Null hypothesis:
`\mu = 634`

Alternative hypothesis:
`\mu \\eq 634`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(669-634)/((732)/(√(1700)))=1.971`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=1700-1=1699`

Since is a two sided test the p value would be:

`p_v =2*P(t_((1699))>1.971)=0.049`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.1` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.