Question

1. Test the hypothesis that the average life of a bulb is 1600 hr. against the alternative that it is 1570 hr. For engineering reasons, only 16 bulbs could be tested obtaining an average of 1590 hr. with a standard deviation of 30 hr. Use a 99% confidence level. Calculate the value of β.

2. An automatic box filling machine has a variability of 0.05 lb. A new automatic machine will be compared with the previous one, and for this, a sample of 18 boxes is taken and the variance is calculated at 0.008 lb. If a 99% confidence level is used, test the hypothesis that the variability remains at 0.05 lb.

3. An oil company states that at most 20% of all car owners buy type A gasoline. Test this hypothesis at the 1% significance level, if a random sample indicates that 58 out of 200 car owners buy type gasoline. TO.

4. A company claims that its lamps are superior to those of its main competitor. If a study showed that a sample of 40 of these lamps has a half-life of 647 hrs. with a standard deviation of 27 hrs., while a sample of 40 of its main competitor had a half-life of 638 hrs. continuous use with a standard deviation of 31 hrs. Should the statement be accepted with a significance level of 0.05?

5. The actual situation of the Fatsa company in terms of the production of carburetors per day, according to the data from the last 8 days, is as follows: 115, 105, 97, 96, 108, 104, 99 and 107. The Model created for the simulation of the plant yields the following 10 carburetor production results per day: 110, 97, 100, 105, 108, 99, 118, 104, 105 and 103. Using a confidence level of 95%, determine if the model results are statistically equal to the real ones. Assume that the production of carburetors follows a normal distribution and that the variances are unknown and equal.

Answer 1

1)

Ho :   µ =   1600                  
Ha :   µ <   1600       (Left tail test)          
                          
Level of Significance ,    α =    0.01                  
sample std dev ,    s =    30.0000                  
Sample Size ,   n =    16                  
Sample Mean,    x̅ =   1590.0000                  
                          
degree of freedom=   DF=n-1=   15                  
                          
Standard Error , SE = s/√n =   30.0000   / √    16   =   7.5000      
t-test statistic= (x̅ - µ )/SE = (   1590.000   -   1600   ) /    7.5000   =   -1.33
                             
                          
p-Value   =   0.1012   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       

2)

Ho :   σ² =   0.05
Ha :   σ² ╪   0.05
      
Level of Significance ,    α =    0.01
sample Variance,   s² =    0.008
Sample Size ,   n =    18
      
Chi-Square Statistic,   X² = (n-1)s²/σ² =    2.72
      
degree of freedom,   DF=n-1 =    17
Two-Tail Test      
Lower Critical Value   =   5.697217101
Upper Critical Value   =   35.71846566
p-Value   =   3.41853E-05
      
Reject the null hypothesis      

3)

Ho :   p =    0.2                  
H1 :   p >   0.2       (Right tail test)          
                          
Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   58                  
Sample Size,   n =    200                  
                          
Sample Proportion ,    p̂ = x/n =    0.2900                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0283                  
Z Test Statistic = ( p̂-p)/SE = (   0.2900   -   0.2   ) /   0.0283   =   3.1820
                          
  
                          
p-Value   =   0.000731358   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                       
There is enough evidence that more than 20% buy gasoline

THANKS

revert back for doubt

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