Many students brag that they have more than 150 friends on a social media website. For a class project, a group of students asked a random sample of 13 students at their college who used the social media website about their number of friends and got the data available below. Is there strong evidence that the mean number of friends for the student population at the college who use the social media website is larger than 150? Required:
a. Find and interpret the test statistic value.
b. Report and interpret the P-value and state the conclusion in context. Use a significance level of 0.05.
c. What does the test statistic value represent?

1. The test statistic value is the difference between the sample mean and the null hypothesis value.
2. The test statistic value is the number of standard errors from the null hypothesis value to the sample mean.
3. The test statistic value is the expected mean of the differences between the sample data and the null hypothesis value.
4. The test statistic value is the number of standard deviations from the null hypothesis value to the sample mean.

Step-by-step explanation:

The question is incomplete. The missing data is:

30, 155, 205, 235, 180, 235, 70, 250, 135, 145, 225, 230, 30

Solution:

Mean = (30 + 155 + 205 + 235 + 180 + 235 + 70 + 250 + 135 + 145 + 225 + 230 + 30)/13 = 163.5

Standard deviation = √(summation(x - mean)²/n

n = 13

Summation(x - mean)² = (30 - 163.5)^2 + (155 - 163.5)^2 + (205 - 163.5)^2+ (235 - 163.5)^2 + (180 - 163.5)^2 + (235 - 163.5)^2 + (70 - 163.5)^2 + (250 - 163.5)^2 + (135 - 163.5)^2 + (145 - 163.5)^2 + (225 - 163.5)^2 + (230 - 163.5)^2 + (30 - 163.5)^2 = 73519.25

Standard deviation = √(73519.25/13) = 75.2

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ ≤ 150

For the alternative hypothesis,

µ > 150

It is a right tailed test.

a) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 13,

Degrees of freedom, df = n - 1 = 13 - 1 = 12

t = (x - µ)/(s/√n)

Where

x = sample mean = 163.5

µ = population mean = 150

s = samples standard deviation = 75.2

t = (163.5 - 150)/(75.2/√13) = 0.65

The lower the test statistic value, the higher the p value and the higher the possibility of accepting the null hypothesis.

b) We would determine the p value using the t test calculator. It becomes

p = 0.26

Since alpha, 0.05 < than the p value, 0.26, then we would fail to reject the null hypothesis. Therefore, At a 5% level of significance, the sample data does not show significant evidence that the mean number of friends for the student population at the college who use the social media website is larger than 150.

c)

1.The test statistic value is the difference between the sample mean and the null hypothesis value.

0.6, 3/5, and 6% are the 3 same portion.

but, when they're all changed to a decimal they would equal to 0.6.

hope this , good luck!

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation: