Please answer each question Show work Thanks in advance Have

Please answer each question. Show work. Thanks in advance!

Have you ever been frustrated because you could not get a container of some product to release the last bit of its contents (make-up, toothpaste, lotion etc)? The article “Shake, rattle and Sqeeze: How Much is Left in That Container?” (Consumers Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose ten 6 oz tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in an average amount remaining of 0.54 oz and sample standard deviation of 0.09 oz. Is there evidence that the true average amount left in the bottle is less than 0.60 oz, that is, less than 10% of the advertised weight? Assume the population to be normally distributed. Use a significance level of 0.05.

State: Is there evidence that the true average amount left in the bottle is less than 0.60 oz, that is, less than 10% of the advertised weight?

Plan:

a. State the null and alternative hypotheses to answer the question of interest.

b. What type of test is appropriate to answer the question of interest and why?

c.State the level of significance.

Solve:

      d.Calculate the test statistic. Xbar=_______ s= _____   n =______

e. What is the p-value for the test? Is it one or two sided?

f.Calculate a 95% confidence interval for µ.

Conclude:

Interpret the results from the hypothesis test and confidence interval in the context of the problem. Use the four step process described at the beginning of the activity to write a conclusion.   

Solution

Formulating the null and alternative hypotheses,          
          
Ho:   u   >=   0.6
Ha:    u   <   0.6 [PART A]

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As we can see, this is a    LEFT   tailed T test, as n = 10 which is quite small.   [PART B]

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level of significance = 0.05 [given, ANSWER, PART C]

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Thus, getting the critical t,          
df =    9      
tcrit =    -1.833112933      
          
Getting the test statistic, as          
          
X = sample mean =    0.54      
uo = hypothesized mean =    0.6      
n = sample size =    10      
s = standard deviation =    0.09      
          
Thus, t = (X - uo) * sqrt(n) / s =    -2.108185107 [ANSWER, PART D]

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Also, the p value is          
          
p =    0.032129425   [ANSWER, PART E, ONE SIDED]

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Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
              
X = sample mean =    0.54          
t(alpha/2) = critical t for the confidence interval =    2.262157163          
s = sample standard deviation =    0.09          
n = sample size =    10          
df = n - 1 =    9          
Thus,              
              
Lower bound =    0.475617878          
Upper bound =    0.604382122          
              
Thus, the confidence interval is              
              
(   0.475617878   ,   0.604382122   ) [answer, part F]

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Comparing z and zcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.      

Thus, there is significant evidence at 0.05 level that that the true average amount left in the bottle is less than 0.60 oz. [CONCLUSION]