I have done part (a) and the histogram is shown. I have determined it is right s

Question

I have done part (a) and the histogram is shown. I have determined it is right skewed and that it makes sense because the mean is 190 and the max is in the 600's. Can someone confirm part (a) and help with parts (b) - (d)? Thanks!

The data from the excel file is the following:

friends 594 60 417 120 132 176 516 319 734 8 31 325 52 63 537 27 368 11 12 190 85 165 288 65 57 81 257 24 297 148 Facebook recently examined all active Facebook users (more than 10% of the global population) and determined that the average user has 190 friends. The Excel file STAT240-Hwk14.P2.Facebook.xls contains data frorn an SRS of n = 30 Facebook Histogranm users from your large university 14 (a) Produce a histogram in Excel of the sample data. What direction is the data skewed? Why does this direction make sense based upon the context of the data? (b) In part (a), you produced a histogram of the sample distribution and observed that it was not Normal. Yet, why do we know that the sampling distribution of the sample mean is approximately Normal? Why is it important to know that 12 10 the sampling distribution of the sample mean is approximately Normal? (c) Conduct a hypothesis test at the 5% significance level to determine if there is sufficient evidence to conclude that the mean number of friends for Facebook users at your large university differs from 190 (the mean for the global population). Be sure to include all steps i. State the hypotheses in proper statistical notation, defining the population parameter in the context of the problem. ii. Compute the test statistic with degrees of m. 0 iii. Report the exact p-value from Excel, as well as an interval for the p-value 200 300 500 700 100 400 600 using the t critical value table iv. State the conclusion in the context of the problem Bin (d) Compute a 95% confidence interval for , the mean number of friends for Face- book users at your large university, and interpret it in the context of the problem

Explanation / Answer

Part b

It is important to know that the sampling distribution of sample mean follows an approximate normal distribution, because we find approximate probabilities for the sample means based on approximate normal distribution.

Part c

Here, we have to use one sample t test for the population mean. The null and alternative hypothesis for this test is given as below:

i)

Null hypothesis: H0: Mean number of facebook friends is 190.

Alternative hypothesis: Ha: Mean number of facebook friends is different from 190.

H0: µ = 190 versus Ha: µ 190

This is a two tailed test.

ii)

The test statistic formula is given as below:

t = (Xbar - µ) / [S/sqrt(n)]

We are given

Sample mean = Xbar = 205.3

Sample standard deviation = S = 196.3312033

Sample size = n = 30

Level of significance = = 0.05

Degrees of freedom = n – 1 = 30 – 1 = 29

Lower critical value = -2.0452

Upper critical value = 2.0452

t = (205.3 – 190) / [196.3312033/sqrt(30)]

t = (205.3 – 190) / 35.8450

t = 0.4268

iii)

P-value = 0.6726

(By using t-table)

iv)

P-value > = 0.05

So, we do not reject the null hypothesis that Mean number of facebook friends is 190.

There is sufficient evidence to conclude that Mean number of facebook friends is 190.

Part d

Confidence interval = Xbar -/+ t*S/sqrt(n)

We are given

Sample mean = Xbar = 205.3

Sample standard deviation = S = 196.3312033

Sample size = n = 30

Level of confidence = 95%

Degrees of freedom = n – 1 = 30 – 1 = 29

Critical t value = 2.0452

Confidence interval = 205.3 -/+ 2.0452*196.3312033/sqrt(30)

Confidence interval = 205.3 -/+ 2.0452* 35.8450096

Confidence interval = 205.3 -/+ 73.3113

Lower limit = 205.3 - 73.3113 = 131.99

Upper limit = 205.3 + 73.3113 = 278.61

Confidence interval = (131, 279)

We are 95% confident that the average number of facebook friends will be lies between 131 and 279.

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