1. An LU professor is interested in whether there is a difference between undergraduate students and graduate students in the amount of time spent praying each day. The professor gathers information from random samples of undergraduate and graduate students on the LU campus. The amount of time praying is normally distributed and is measured on an interval/ratio scale.
Graduate Undergraduate
15 9
17 11
10 9
13 6
11 5
17 6
a. What statistical test should be used to analyze the data?
b. Is this a one- or two tailed test?
c. Identify H0 and Ha for this study.
d. Conduct the appropriate analysis. Should H0 be rejected?
a. The appropriate test for analyzing this data would be the independent samples t-test. This is because we are comparing the means of two independent groups (graduate students and undergraduate students) to see if there is a significant difference in the amount of time spent praying.
b. This is a two-tailed test. We are interested in determining whether there is a significant difference in the amount of time spent praying between the two groups, so we want to consider both possibilities (that the graduate students pray more and that the undergraduate students pray more).
c. H0: There is no difference in the amount of time spent praying between undergraduate and graduate students.
Ha: There is a difference in the amount of time spent praying between undergraduate and graduate students.
d. To conduct the independent samples t-test, we compare the means of the two groups using the following steps:
1. Calculate the mean and standard deviation of the amount of time spent praying for both the graduate and undergraduate groups.
Graduate:
Mean = (15 + 17 + 10 + 13 + 11 + 17) / 6 = 83 / 6 = 13.83
Standard deviation = 2.72
Undergraduate:
Mean = (9 + 11 + 9 + 6 + 5 + 6) / 6 = 46 / 6 = 7.67
Standard deviation = 2.13
2. Calculate the t-statistic:
t = (mean_graduate - mean_undergraduate) / sqrt((squared_std_graduate / n_graduate) + (squared_std_undergraduate / n_undergraduate))
t = (13.83 - 7.67) / sqrt((2.72^2 / 6) + (2.13^2 / 6))
t = 6.16 / sqrt((7.41 / 6) + (4.53 / 6))
t = 6.16 / sqrt(1.234 + 0.755)
t = 6.16 / sqrt(1.989)
t = 6.16 / 1.41
t = 4.37
3. Look up the critical value for a two-tailed test with the desired alpha level (typically 0.05). In this case, with a sample size of 6, the critical t-value is approximately 2.571.
4. Compare the calculated t-value (4.37) with the critical t-value (2.571). If the calculated t-value is greater than the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the calculated t-value is greater than the critical t-value, so we reject the null hypothesis. There is sufficient evidence to suggest that there is a significant difference in the amount of time spent praying between undergraduate and graduate students.
a. The appropriate statistical test to analyze the data in this case is an independent samples t-test.
b. This is a two-tailed test since we are interested in whether there is a difference between undergraduate and graduate students in the amount of time spent praying, without specifying a particular direction of difference.
c. H0 (null hypothesis): There is no difference in the amount of time spent praying between undergraduate and graduate students.
Ha (alternative hypothesis): There is a difference in the amount of time spent praying between undergraduate and graduate students.
d. To conduct the analysis, we will use a statistical software or calculator. The independent samples t-test will compare the means of the two groups (undergraduate and graduate) and determine if there is a statistically significant difference between them.
Let's calculate the t-value and p-value using the given data:
Undergraduate students: 9, 11, 9, 6, 5, 6
Graduate students: 15, 17, 10, 13, 11, 17
The mean for undergraduate students (x̄1) is (9+11+9+6+5+6)/6 = 46/6 = 7.67
The mean for graduate students (x̄2) is (15+17+10+13+11+17)/6 = 83/6 = 13.83
The standard deviation for undergraduate students (s1) can be calculated as:
√[((9-7.67)^2 + (11-7.67)^2 + (9-7.67)^2 + (6-7.67)^2 + (5-7.67)^2 + (6-7.67)^2) / 5] = √[11.17+2.44+11.17+2.44+6.11+2.44 / 5] = √6.3157 = 2.51
The standard deviation for graduate students (s2) can be calculated as:
√[((15-13.83)^2 + (17-13.83)^2 + (10-13.83)^2 + (13-13.83)^2 + (11-13.83)^2 + (17-13.83)^2) / 5] = √[5.32+8.67+11.16+0.68+5.77+8.67 / 5] = √7.074 = 2.66
The t-value can be calculated as:
t = (x̄1 - x̄2) / √((s1^2/n1) + (s2^2/n2)) = (7.67 - 13.83) / √((2.51^2/6) + (2.66^2/6)) = -6.16 / √(0.4192 + 0.4696) = -6.16 / √0.8888 = -6.16 / 0.9428 = -6.54
Degrees of freedom (df) can be calculated as:
df = n1 + n2 - 2 = 6 + 6 - 2 = 10
Now, we need to look up the critical t-value for a two-tailed test with 10 degrees of freedom and a significance level of 0.05. Looking this up on a t-table, we find the critical t-value to be approximately ±2.228.
Since the calculated t-value (-6.54) is less than the critical t-value (-2.228), we can reject the null hypothesis (H0). There is strong evidence to suggest that there is a significant difference in the amount of time spent praying between undergraduate and graduate students on the LU campus.