Hello, Can I please get some help with this question? (a,b,c,d,e,f,g) I need to
ID: 3228776 • Letter: H
Question
Hello, Can I please get some help with this question? (a,b,c,d,e,f,g) I need to be able to show all the steps so I can better understand how to get the answer for each. Thank you in advance!
4) In addressing the same question, let's assume you take five statistics students and give them a lesson on a sunny day, and then take the same five students and give them a different statistics lesson on a rainy day. These are their results for quizzes on these lessons.
Sunny Day Results:
Student 1: 98
Student 2: 67
Student 3: 72
Student 4: 82
Student 5: 51
Rainy Day Results:
Student 1: 82
Student 2: 67
Student 3: 68
Student 4: 97
Student 5: 71
a.State the null and alternative hypotheses
b. What are the degrees of freedom for your t test? Find the corresponding critical t-value for Type I error rate (alpha) of = 0.05
. c. Calculate your observed t-statistic (hint: you will need to calculate the average difference scores and the SD of the average difference score first).
d. Compare your observed t-statistic to the critical t-value(s). What do you conclude regarding the null hypothesis?
e. Calculate and interpret the 95% Confidence interval.
f. Calculate and interpret the standardized effect size (Cohen's d).
g. What do you conclude about your research question (use your own words, in everyday language)
Explanation / Answer
a.State the null and alternative hypotheses
Null Hypothesis : H0 : There is no difference in quiz scores for students whethere they take lecture on rainy day or sunny day.d = 0
ALternative Hypothesis : Ha : There is significant differencein quiz scores for students when they lecture on rainy or sunny day. d 0.
b. What are the degrees of freedom for your t test? Find the corresponding critical t-value for Type I error rate (alpha) of = 0.05
Degrees of freedom = n -1 = 4 as it is paired data n would be 5 only.
For alpha = 0.05 and dF = 4
tcritical = 2.7764 ( you can see this value in t - table)
. c. Calculate your observed t-statistic (hint: you will need to calculate the average difference scores and the SD of the average difference score first).
Test Statistic : I am presenting difference table here
so Test Statistic:
t = dbar / (Sd/ n) = 3/ (14.59/ 5) = 3/ 6.5248 = 0.46
d. Compare your observed t-statistic to the critical t-value(s). What do you conclude regarding the null hypothesis?
Here t < tcrtical so we can not reject the null Hypothesis.
e. Calculate and interpret the 95% Confidence interval.
95 % confidence interval. = dbar +- Z0.05 ((Sd/ n) = 3 +- 1.96 (6.5248) = (-9.79, 15.79)
f. Calculate and interpret the standardized effect size (Cohen's d).
effect size = t/ n = 0.46/ 5 = 0.205
Effect size is small in nature
g. What do you conclude about your research question (use your own words, in everyday language)
We can conclude that there is no effect of weather or environment on the marks of students for a perticulat quiz.
sunny Rainy d 98 82 -16 67 67 0 72 68 -4 82 97 15 51 71 20 Average 3 Std. dev. 14.59Related Questions
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