Performing T Tests to Calculate Means

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 of 3. 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
b) Create the appropriate null and alternate hypothesis for this test. 
H0 = younger adults use social media more than adults aged 35 or over
H1 = younger adults do not use social media more than adults aged 35 or over
Part 2
1. A random sample of 25 U.S. adults were asked to estimate the average income of all U.S. households. The estimate mean was $47,000 with a standard deviation of $15,000. Construct a 95% confidence interval for this data
95% confidence interval for the population mean is computed as follows:
X ± 1.96 ? / ?N
= 47,000 ± 1.96 15,000 / ?25
= 47,000 ± 5,880
= (41,120, 52,880)
2. People claim that women say more words per day than men. Estimates claim that a woman uses roughly 20,000 words per day, while a man uses approximately 7,000. To investigate this, a researcher recorded conversations of male college students over a 5-day period. The results are as follows:
            7220                       13932                    4727                       10419
            9258                        9717                    10728                       5265
           12215                       9944                     7979                       12252
            9307                        9086                    10780                       3357 
The researcher believes that many use more than 7,000 words per day.
a) State the problem in your own words.
The problem is to make a determination as to who says more words between men and women
b) Create a plan for testing the researcher’s claim. Be sure to state the null and alternate hypotheses.
H0 = women say more words than men
H1 = women do not say more words than men
c) Carry out the appropriate hypothesis test. Begin by finding the sample mean and standard deviation. Give the value of the t statistic and give the p-value (or an estimate).
Sample mean = summation of the conversations divided by the number of students
= 146186 / 16
= 17.198.35
Standard deviation = 33,354.74
T test
x? = sample mean = 7000
?0 = population mean = 9136.63
s = sample standard deviation = 31,354.74
n = sample size = 16
t = (x? - µ0) / (s /?n)
= (9136.62 - 7000) / (31,354.74 /?16)
= 2136.62 / 7838.69
= 0.2726
d) Formulate the statistical conclusion in terms of the null hypothesis and a practical conclusion. You may compare the p-value to 0.05 to determine if the null hypothesis should be rejected. When the p-value is less than 0.05, we reject the null hypothesis, otherwise we fail to reject the null.
The null hypothesis should be accepted
3. A manufacturer states that the average lifetime of its light bulbs is 3 years. A random sample of 50 bulbs is taken and the average lifetime is found to be 34 months. The population is normal with a standard deviation of 8 months. Should the claim be rejected using the 0.05 as a benchmark for the p-value? Construct the hypothesis test for this using the four-step process. Write a detailed description of the steps: state, plan, solve, and conclude.
State
Null hypothesis: average lifetime of light bulbs is 3 years
Alternative hypothesis: average lifetime of light bulbs is not 3 years
Plan
Sample Mean = 34 months
Standard deviation = 8
Population mean = 36 months
Sample size = 50
Solve
T- test computed as:
(34 – 36) / (8 / ?50)
= -1.7678
The P-Value is .041728.
Conclude
Accept null hypothesis
Therefore, average lifetime of light bulbs is 3 years (Mendenhall et al., 2012)
4. A marketing research firm wants to test whether the mean rating given by consumers to their favorite beer (denoted by X) differs from some other beer (denoted by Y). The ratings obtained from two different random groups are given below. Perform a t test to determine if there is a significant different in the mean ratings of these two beers. Use Excel to perform the calculations. See the website below for instructions.
http://www.excel-easy.com/examples/t-test.html
When explaining the test and its results, write a detailed description of the steps: state, plan, solve, and conclude. Assume unequal variances, and use 0.05 as a benchmark for the p-value. 
          Brand X                                   Brand Y
                57                                           63
                61                                           60
                62                                           60
                60                                           62
                60                                           63
                62                                           59
                58                                           57
                57                                           64
                62                                           62
 State:
Null hypothesis: the mean rating given by consumers to their favorite beer X differs from Y
Alternative hypothesis: the mean rating given by consumers to their favorite beer X does not differ from Y
Solve: Find t test
=T.TEST(E1:E9, F1:F9,2,1)
= 0.329009236

Conclusion: Reject the null hypothesis
5. Create mock scenario and mock data for a matched pairs t test. The data set should include at least 10 pairs of data. Perform a matched pairs t test using the four-step process described above. Use Excel to perform the calculations. See the website below for instructions:
http://www.real-statistics.com/students-t-distribution/paired-sample-t-test/

State
Class A performed better than Class B
Data for 10 students for test is as follows:
Class A Class B
84 90
89 94
90 88
92 96
88 82
82 95
94 89
98 88
86 92
90 80
Plan
Null hypothesis: Class A performed better than Class B
Alternative hypothesis: Class A performed worse than Class B
Solve
T-test is: =T.TEST(H1:H10,I1:I10,2,1)
= 0.968527173
Conclude
The P-Value is .344963.
The result is not significant at p < .05.
Therefore, accept null hypothesis




References
Ríos, V. R., & del Campo, E. P. (2013). Business research methods: theory and practice. ESIC editorial.
Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2012). Introduction to probability and statistics. New York: Cengage Learning.