Performing t-tests for means
Part 1
1. Explain how the t distribution is similar to a normal distribution, and how it is different from a normal distribution
There are a number of similarities between the t distribution and the normal z distribution. To begin with, both of these distributions are symmetric, in the sense that they have a bell-like shape, about a mean of zero. In this regard, the t distribution can be utilized as a substitute to the normal distribution in the instances where the sample sizes are small so as to make an estimation of the confidence. However, at the same time, there are dissimilarities between the two distributions. First of all, the t distribution is in fact more variable as compared to the normal distribution. This implies that it is spread out and therefore has a smaller peak and end tails that are fatter. Secondly, whilst there is solely one normal z distribution, there exists several t distributions. With the increase in the number of degrees of freedom, the t distribution approaches the z distribution. Taking this into consideration, in the event of large samples, the normal distribution and the t distribution are almost undistinguishable (Dios and del Campo, 2013).
2. Explain the differences in terms of the null hypothesis between a one-sample t test, a two sample t test, and a matched pairs t test
The null hypothesis in a single sample t-test assumes that no difference exists amid the comparison and true mean. The null hypothesis in a two sample t-test assumes the true value is greater or smaller than the comparison mean. The null hypothesis in a matched pairs t-test assumes that no difference exists between the comparison and the true mean and the for different populations.
3. Suppose that you are testing the null hypothesis H0: µ = 100 against Ha: µ < 100 based on a random sample of nine observations from a normal population. The data from the sample give a mean of 98 and a standard deviation...
What is the value of the t statistic?
t = (x? - µ0) / (s - ?n)
In this case:
x? = sample mean = 98
?0 = population mean = 100
s = sample standard deviation = 3
n = sample size = 9
t = (x? - µ0) / (s /?n)
= (98 - 100) / (3 /?9)
= -2 / 1
= -2
4. In Problem 3, either find the exact p-value or determine a lower and upper bound of the p-value for the test statistic
The P-Value is 0.040258.
5. Identify the correct type of t test to use in the following situations. Choose from a 1-sample, 2-sample, or matched pairs t test.
a) You interview 500 female students and ask each about the average number of minutes they spend using the internet each day.
1 sample t test
b) You interview a sample of 250 unmarried male students and 250 unmarried female students and ask each about the average number of minutes they spend using the internet each day
2 sample t test
c) You interview 200 female students in their freshman year and again in their senior year and ask each about the average number of minutes they spend using the Internet each day
Matched pairs
6. Explain how the degrees of freedom are determined using:
a) a one sample t test
For a one sample t test, the degrees of freedom are determined by n -1
b) a two sample t test
For a two sample t test, the degrees of freedom are determined by the one that has a smaller number between n1 – 1 and n2 - 1
7. A marketing company believes that younger adults use social media more than adults aged 35 or over. To investigate this, the company is planning to take a random sample of each group and determine the sample means.
a) Determine which type of statistical test is appropriate.
The appropriate test is the 1 sample t test…
