The population that is your PSYC 200 class has been polled as to their adaptabil
ID: 3296016 • Letter: T
Question
The population that is your PSYC 200 class has been polled as to their adaptability. Adaptability was measured using 4 items, which addressed issues relevant to taking advice, adapting to new situations, tolerating criticism, and having the last word. A “1" on this scale would mean “low adaptability". A "5" on this scale would mean "high adaptability" You want to see if your study group from class is different from the population. The mean score on adaptability for this population is 3.2 and the standard deviation is 1 Suddenly, you realize that what you thought was the standard deviation was in fact a typo and you do not actually know the standard deviation of the population! Using the same scores as you drew in your first problem: 3.5, 4.75, 3.5, 3.5, 4, 5 Conduct a formal null hypothesis significance test, making sure to include every step and to show all of your work, making sure to state the hypotheses in words and using statistical notation.Explanation / Answer
State the hypotheses: The null and alternative hypotheses are as follows:
H0:mu=3.2 (mean score on adaptibility of the study group from class is no different from the population)
H1:mu=/=3.2 (mean score on adaptibility of the study group from class is different from the population)
Assumptions: Assume the study group were sampled in randomized way, and the adaptability score of the population is normally distributed. The conditions being met, usr Student's t model with (n-1)=(6-1)=5 degrees of freedom.
Test statistic.
t=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard deviation, and n is sample size. Using calculator, xbar=4.042, s=0.679.
=(4.042-3.2)/(0.679/sqrt 6)
=3.04
p value at 5 df is 0.029.
Rejection criteria: Rejection rule states that reject null hypothesis if p value is less than alpha=0.05. Here, p value is less than 0.05, thus, reject H0.
Conclusion: Conclude that mean score on adaptibility of the study group from class is significantly different from the population.
The 95% c.i for population mean adpatability score, mu=xbar+-talpha/2, df=n-1 (s/sqrt n) , where, talpha/2 is t critical at alpha/2 (alpha=0.05) and 5 df.
=4.042+-2.5706(0.679/sqrt 6)
=(3.33, 4.75) [width:Upper bound-Lower bound=4.75-3.33=1.42]
Once, the interval is decresed from 95% to 90%, the margin of error [ME=talpha/2, df=n-1(s/sqrt n)] will decrease resulting into smaller width.
The 90% c.i for mu is: 4.042+-2.015(0.679/sqrt 6)
=(3.48, 4.60) [width:1.12]
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