× Didn't find what you were looking for? Ask a question
  
  
Top Posters
Since Sunday
26
21
21
21
20
19
19
19
18
18
18
18
New Topic  
wrote...
Posts: 1
Rep: 0 0
3 weeks ago
6.   A psychology professor of a large class became curious as to whether the students who turned in tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 students to turn in tests was 78. Using the .05 significance level, was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?

a.   Use the five steps of hypothesis testing.
Read 81 times
3 Replies
Related Topics
Replies
wrote...
3 weeks ago
Hi brendalee

Null hypothesis: Mean test score for students who turned in the test first = overall test mean

Alternative hypothesis: Mean test score for students who turned in the test first ≠ overall test mean

Significance level: .05

Test statistic / critical value: t-value of 1.96 (note that this critical value is given for a sample of infinite size. Because n of the overall sample has not been provided, t cannot be precisely calculated)

Conclusion: Reject the null hypothesis if the |t| of the comparison between the two means < 1.96

Confidence limits for the 95% confidence interval

Assuming n = 100, the 95% confidence interval is given by the formula of 75 ± (2.04)(1.8) = 71.33 to 78.67

[Note that the 95% confidence interval for the entire sample cannot be perfectly estimated because n is not given, and the 95% confidence interval for the students who turned in their test first cannot be calculated because the standard deviation for this sample has not been given].
wrote...
3 weeks ago
The provided sample mean is 78, and the known population standard deviation is σ=10, and the sample size is n = 20

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ  = 75

Ha: μ≠75

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=.05, and the critical value for a two-tailed test is z = 1.96

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

z = 1.342

(4) Decision about the null hypothesis

Since it is observed that ∣z∣=1.342≤z =1.96 , it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.1797, and since p=0.1797≥.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean

μ is different than 75, at the .05 significance level.

Confidence Interval

The 95% confidence interval is given as 73.617<μ<82.383.
wrote...
Educator
3 weeks ago
In this post, https://biology-forums.com/index.php?topic=139802.0, A and D are answered and accepted as best answer too!
The best way to say thank you is with a positive review:

  https://trustpilot.com/review/biology-forums.com 

Your support goes a long way!


Make a note request here
New Topic      
Explore
Post your homework questions and get free online help from our incredible volunteers.
Learn More
Improve Grades
Help Others
Save Time
Accessible 24/7
  102 People Browsing
Related Images
 708
 147
 71