Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

An experimenter is interested in the hypothesis testing problem H_0: mu = 430.0

ID: 3153885 • Letter: A

Question

An experimenter is interested in the hypothesis testing problem H_0: mu = 430.0 versus H_A: mu notequalto 430.0 where mu is the average breaking strength of a bundle of wool fibers. Suppose that a sample of n = 20 wool fiber bundles is obtained and their breaking strengths are measured. For what values of the t-statistic does the experimenter accept the null hypothesis with a size alpha = 0.10? For what values of the t-statistic does the experimenter reject the null hypothesis with a size alpha = 0.01? Suppose that the sample mean is x = 436.5 and the sample standard deviation is s = 11.90. Is the null hypothesis accepted or rejected with alpha = 0.10? With alpha = 0.01? Write down an expression for the p-value and evaluate it using a computer package.

Explanation / Answer

a)

As df = n - 1 = 19, then for two tailed, 0.10 level, using table/technology, the non-rejection region is

-1.729132812 < t < 1.729132812 [ANSWER]

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

B)

As df = n - 1 = 19, then for two tailed, 0.01 level, using table/technology, the rejection region is

t < -2.860934606 OR t > 2.860934606 [ANSWER]

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

c)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   430  
Ha:    u   =/   430  
              
As we can see, this is a    two   tailed test.      
              
df = n - 1 =    19          
              
Getting the test statistic, as              
              
X = sample mean =    436.5          
uo = hypothesized mean =    430          
n = sample size =    20          
s = standard deviation =    11.9          
              
Thus, t = (X - uo) * sqrt(n) / s =    2.442763337          
Using part a), we REJECT Ho at 0.10 level. [ANSWER]

Using part b), we ACCEPT HO at 0.01 level. [ANSWER]

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

d)
              
Also, the p value is, using table,

0.02 < P < 0.05 [ANSWER]

Using technology,              
              
p =    0.024517374   [ANSWER]      
              

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote