Will someone please help me with this statistics problem?

Question

For each of the situations below, chose the correct test, and then use the data to conduuct your chosen test. Afterward, explain why it was the appropriate test and what the results were. Also, explaning whether or not a significant difference was found between the sample and the population.

A researcher wants to compare the email use of the employees at one company to the corresponding population. Data is collected on the number of emails received by the company's employees. It will be necessary to use this sample data, as well as population data to determine if there is a significant difference between both.

Population

1. Mean number of emails: 30
2. Standard deviation: 4

Sample

1. Mean number of emails: 33
2. Total employees: 100

A researcher wants to compare the number of vehicles owned by families in one location to the larger population. Data is collected on the number of vehicles owned by each family. It will be necessary to use this sample data, as well as population data to determine if there is a significant difference between them.

Population

1. Mean number of vehicles: 3.7

Sample

1. Mean number of vehicles: 4.1
2. Standard deviation: 1
3. Sample size: 100

Answer 1

Hi Sarah,

#1. You know population standard deviation sigma, and n is sufficiently large, so you can use a two-sided z-test. (It could also be done as a t, but significance decision will not matter) Formula for test statistic is:

z= (xbar-mu)/SE

xbar= sample mean = 33

mu= population mean = 30

SE=standard error= sigma/sqrt(n)

SE= 4/sqrt(100)

SE= 0.4

Now, plug those values into the equation above to get your z-test statistic:

z= (xbar-mu)/SE

Then, you can either look that statistic up in the z-table or type into a TI-80s series calculator (hint: it may not appear in z-table; what does that tell you?):

In calculator:

STAT-TESTS-Z-TEST-Input mu, sigma, xbar, and n where directed. Regardless of method, your p-value will be very small. I will leave the interpretation of significance to you based on that.

#2.

Here, you must use the one-sample t-test, as you do not know a population standard deviation. T-test statistic formula is similar. It is still:

t= (xbar-mu)/SE; BUT

SE=s/sqrt(n)

Note we replace population standard deviation sigma with sample standard deviation s.

xbar= 4.1

s=1

n=sample size= 100

SE= (1/sqrt(100))

SE=0.1

Now, we plug those values into the equation above:

t= (xbar-mu)/SE

You will then have your t-test statistic, but there is one more step required. For t-tests, we also need degrees of freedom (df). Formula is simple:

df= n-1

df= 100-1 = 99

Now, look to a t-table, you will not see 99 degrees of freedom listed, so you will have to measure based on the next closest without going over; usually 80. Again, you can check the t-table for your test statistic or use the TI-80s series:

STAT- TESTS - T-Test - Input:Stats- enter mu, xbar, and s_x where requested, calculate. Again, either way, your p-value will be very small. I will leave you to interpret significance based on that.

I hope this helps.