Answer:
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.
