Will Someone Please Help Me With This Statistics Problem.

Question

Consider the following: Analysis of national data have suggested that the adult population uses social media an average of 2.4 hours per day, with a standard deviation of 1.4 hours. The sample mean of a group of 10 people is 2.9. Address the following.

Use the sample mean, along with the population mean and standard deviation provided above, to conduct a one-sample z-test. Give the results of the z-test.

Explain whether or not the difference between your sample and the population is statistically significant.

Discuss some reasons for this significant difference or lack thereof. 

Answer 1

Hi Sarah,

Formula for your test-statistic, z, is:

z= (xbar - mu)/SE

xbar=sample mean

mu=population mean

SE=standard error

SE=sigma/sqrt(n)

sigma= population standard deviation

n=sample size

For our problem:

Sigma= 1.4

n = 10

SE= 1.4/sqrt(10)

SE= 0.443

Now, we plug that value into our z-test statistic formula:

z= (xbar-mu)/SE

xbar = 2.9

mu = 2.4

SE = 0.443

z= (2.9-2.4)/0.443

z= 1.13

Now, your question did not specify which direction you think this leans, i.e. do we believe adults use social media more than, less than, or simply not equal to, 2.4 hours per day? The question did not specify, so I will use the not equal to alternative. For this alternative, you must look to the z-score table, find:

1- P(Z<1.13) and multiply by 2. Don't forget this last step as "not equal to" means two-sided, which means multiplying by 2.

P(Z<1.13)= 0.8708

1 - P(Z<1.13) = 1- 0.8708 = 0.1292

p=2(0.1292)= 0.2584

Again, the question did not specify a key piece of information, the significance level alpha, but the most common is 0.05, so we will use that here. If p<0.05, result is statistically significant. If not, result is not statistically significant. I will leave the judgment and reasons to you, but think about sample size and variability within the population--"adults" encompass a wide range of socioeconomic and occupational strata. I hope this helps.