Ms. Fitness-Buff, a high school gym teacher, wants to propose an after school fi

Question

Ms. Fitness-Buff, a high school gym teacher, wants to propose an after school fitness program. To get an idea of the fitness level of the student at her school, she takes a random sample of 75 students and records the number of hours the students exercised in the past week. Her sample mean in 2.25 hours and she knows from past research that the population standard deviation is 2 hours. She wants to know if this varies from a population mean of 3 hours per week.

Construct a 95% confidence interval.

Draw a conclusion for a two-sided test at a = .05.

Ms. Fitness-Buff is told that if students exercise less than 3 hours per week, she can start an after-school fitness program. Test this one-sided hypothesis and draw a conclusion at a = .05.

What would the consequences of a Type I error be in the test from part C?

What's the probability of a Type I error for the test in part C?

What would the consequences of a Type II error be for the test from part C?

What is the rejection region for Ho: mue = 3 for the test from part C?

Calculate the probability of making a Type II error if the true population mean is 2.75.

What's the power of the test if the true population mean is 2.75?

Explanation / Answer

A)

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    2.25          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    2          
n = sample size =    75          
              
Thus,              
              
Lower bound =    1.797365706          
Upper bound =    2.702634294          
              
Thus, the confidence interval is              
              
(   1.797365706   ,   2.702634294   ) [ANSWER, CONFIDENCE INTERVAL]

***************************

b)

Thus, at 0.05 level, there is significant evidence that the exercise time varies from 3 hrs/week, as this confidence interval does not contain 3.

*****************************

c)

Ms. Fitness-Buff is told that if students exercise less than 3 hours per week, she can start an after-school fitness program. Test this one-sided hypothesis and draw a conclusion at a = .05.

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   3  
Ha:    u   <   3  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    2.25          
uo = hypothesized mean =    3          
n = sample size =    75          
s = standard deviation =    2          
              
Thus, z = (X - uo) * sqrt(n) / s =    -3.247595264          
              
Also, the p value is              
              
p =    0.000581923          
              
Comparing z and zcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.          
              
Thus, there is significant evidence at 0.05 level that the mean exercise time per week is less than 3 hours. [CONCLUSION]

******************************

d)

A type 1 error is incorrectly saying that the mean weekly exercise time is less than 3 hours, when in fact, it is not.

****************************

e)

The probability of a type I error is the significance level, 0.05. [answer]

*******************************************

Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.