Steps to be covered state the hypothesis and identify the cl

Steps to be covered:

state the hypothesis and identify the claim

find the critical value from the table and mention the acceptance range

compute the test value

make the decision to reject or not the null hypothesis

Note:   We will consider all CIE as two-tailed test(s);

Claim tests are two-tail tests only if you have to test for strict equality;

If the claim says “at least”, “less”, “at most”, “no more than”….. it means that a one-tail test should be performed.

(>, <, >= , <= are all one-tail test)

A#

H&M store claims that its batteries have a life of more than 600 hours. The Mall Inspectors test a sample of 20 batteries, and found:

= 595 hours

s = 40 hours

Test at a=.05

Construct a 2-tailed 95% C.I.E of . The upper limit of the confidence interval is ______________

#B

A doctor claims that the average number of infections per week at a hospital in Westchester is 16.3. A random sample of 10 weeks had a mean number of 17.7 infections. The sample standard deviation is 1.8. Is there enough evidence to reject the investigator’s claim at a=.05?

n = 10

s = 1.8

= 17.7

Solution

A.

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   600  
Ha:    u   <   600  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    19          
tcrit = -1.729132812      
              
Getting the test statistic, as              
              
X = sample mean =    595          
uo = hypothesized mean =    600          
n = sample size =    20          
s = standard deviation =    40          
              
Thus, t = (X - uo) * sqrt(n) / s =    -0.559016994          
              
Also, the p value is              
              
p =    0.291339917          
              
Comparing |t| < 1.729, (or P > 0.05), we   FAIL TO REJECT THE NULL HYPOTHESIS.          

There is no significant evidence that the mean life is not more than 600 hours. [CONCLUSION]

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Note that              
      
      
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    595          
t(alpha/2) = critical t for the confidence interval =    2.093024054          
s = sample standard deviation =    40          
n = sample size =    20          
df = n - 1 =    19          
Thus,              

Upper bound =    613.7205763   [ANSWER]      
              

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