The weights of the apples grown on an orchard are normally distributed. the mean weight has been m0 = 9.500 ounces. because of the inclement weather this year, the farmer would like to know if the mean weight has decreased. a random sample of n = 16 apples is selected. the sample mean is x = 9.320 ounces and the sample standard deviation is s = 0.18 ounces. a test needs to be conducted to test if the population mean has decreased at a significance level a = 0.05 .

23. the null and alternative hypotheses should be (3)

24. using the action limits to set up the decision rule, the decision rule should be (3)

25. the test statistic of this test is (3)

26. the conclusion of this test should be (3)

t(s) is in rejection zone then we reject H₀.

Bad weather indeed make apples weight decrease

Step-by-step explanation:

Normal Distribution

population mean    μ₀   = 9.5 ou

sample size   =  n   =  16  then we should apply t-student table

degree of fredom   df  =  n  - 1     df =  16  -  1    df  = 15

1.-Test  hypothesis

H₀     null  hypothesis                          μ₀   =  9.5    

Hₐ alternative hypothesis                   μ₀  <  9.5

One left tail-test

2.-Confidence level 95 %

α  =  0,05    and   df  = 15    from t-student table we get  t(c)  =  - 1.761

3.-Compute t(s)  

t(s)  =  [ μ  -  μ₀  ] /√s/n         t(s)  = (9.32  -  9.5 )* √16 / 0.18

t(s)  = -  0.18*√16 / 0.18

t(s)  = - 4

4.-Compare  t(s)   and t(c)

t(s) < t(c)     -4  <  - 1.761

Then  t(s)  is in the rejection zone.

5.- Decision

t(s) is in rejection zone then we reject H₀.

Farmer conclude that bad weather make apples weight decrease

mr. graef told us it is 13 pints

step-by-step explanation:

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square