A study investigated the job satisfaction of teachers allowed to choose supplementary curriculum for their classes versus teachers who were assigned all curricular resources for use in their classes. On average, when surveyed regarding job satisfaction, teachers give a score of 3.3 out of 5 with a standard deviation of 0.6. When the authors of the study interviewed 40 teachers who supplemented with their own materials, they found 3.5 to be the mean. The authors wanted to know if the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction. They used a significance level of 1%. Which of the following statements is valid based on the results of the test?
What statements?
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
nnmnm
To determine if the group of teachers that could choose supplementary curriculum had a higher level of job satisfaction, a hypothesis test needs to be conducted. Based on the information provided, we can set up the following hypotheses:
Null Hypothesis (H0): The mean job satisfaction of teachers who could choose supplementary curriculum is equal to or less than the mean job satisfaction of teachers who were assigned all curricular resources.
Alternate Hypothesis (H1): The mean job satisfaction of teachers who could choose supplementary curriculum is higher than the mean job satisfaction of teachers who were assigned all curricular resources.
To test this hypothesis, we can use a one-sample t-test. The formula for the t-test statistic is:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
Given the information provided, the sample mean is 3.5, the population mean is 3.3, the sample standard deviation is 0.6, and the sample size is 40.
Calculating the t-test statistic:
t = (3.5 - 3.3) / (0.6 / sqrt(40))
t = 0.2 / (0.6 / sqrt(40))
t = 0.2 / (0.6 / 6.324)
t = 0.2 / 1
t = 0.2
To determine if the t-test statistic is statistically significant, we need to compare it to the critical value from the t-distribution at the given significance level of 1%.
Looking up the critical value in a t-table or using statistical software, we find that the critical value for a one-tailed test at 1% significance level with 39 degrees of freedom (40-1) is approximately 2.432.
Since the calculated t-value of 0.2 is less than the critical value of 2.432, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the group of teachers who could choose supplementary curriculum had a higher level of job satisfaction compared to the group of teachers who were assigned all curricular resources.
Hence, based on the results of the test, the valid statement is that there is no statistically significant difference in job satisfaction between the two groups of teachers.