Respuesta :
Answer:
No, the sample does not provide evidence that the average number of social ties for a cell phone user is significantly different from 634.
Step-by-step explanation:
In this case we need to determine whether the average number of social ties for a cell phone user is significantly different from 634 people.
The hypothesis can be defined as follows:
H₀: The average number of social ties for a cell phone user is 634 people, i.e. μ = 634.
Hₐ: The average number of social ties for a cell phone user is different from 634 people, i.e. μ ≠ 634.
The information provided is:
[tex]n=1700\\\bar x=664\\SD=778[/tex]
As the sample size is quite large, i.e. n = 1700 > 30 and is taken form an unknown population, according to the Central limit theorem the sampling distribution of sample mean will follow the Normal distribution.
So, a z-test for single mean will be applied to perform the test.
Compute the test statistic value as follows:
[tex]z=\frac{\bar x-\mu}{SD/\sqrt{n}}\\\\=\frac{664-634}{778/\sqrt{1700}}\\\\=1.59[/tex]
The test statistic value is 1.59.
Decision rule:
If the p-value of the test is less than the significance level then the null hypothesis will be rejected.
Compute the p-value for the two-tailed test as follows:
[tex]p-value=2\cdot P(Z >1.59)\\=2\times [1-P(Z<1.59)]\\=2\times (1-0.944)\\=0.112[/tex]
*Use a z-table for the probability.
The p-value of the test is 0.112.
The p-value of the test is very large for all the commonly used significance level. The null hypothesis will not be rejected.
Thus, there is not enough evidence to conclude that the average number of social ties for a cell phone user is different from 634 people.
