Question

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was 9 months (3/4 year) with a standard deviation of 3 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old. Conduct a hypothesis test at the 5% level.

Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)

  State the null hypothesis.

  H₀:

  μ ≤ 4

  H₀:

  μ = 4     

  H₀:

  μ ≠ 4

  H₀:

  μ > 4

• Part (b)

  State the alternative hypothesis.

  H_a:

  μ < 4

  H_a:

  μ ≠ 4     

  H_a:

  μ = 4

  H_a:

  μ > 4

• Part (c)

  In words, state what your random variable

  X

  represents.

  X represents the age at which American mothers wean their children.

  X represents the number of children weaned by age 4.

       

  X represents the average number of children weaned by age 4.

  X represents the average age at which American mothers wean their children.

• Part (d)

  State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom.)

• Part (e)

  What is the test statistic? (If using the z distribution round your answers to two decimal places, and if using the t distribution round your answers to three decimal places.)
  ---Select--- t z =

• Part (f)

  What is the p-value? (Round your answer to four decimal places.)
  
  
  Explain what the p-value means for this problem.

  If H₀ is false, then there is a chance equal to the p-value that the average age at which American mothers wean their children is equal to the sample mean or more.If H₀ is true, then there is a chance equal to the p-value that the average age at which American mothers wean their children is equal to the sample mean or less.     If H₀ is false, then there is a chance equal to the p-value that the average age at which American mothers wean their children is equal to the sample mean or less.If H₀ is true, then there is a chance equal to the p-value that the average age at which American mothers wean their children is equal to the sample mean or more.

• Part (g)

  Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)

  Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.(i) Alpha (Enter an exact number as an integer, fraction, or decimal.)
  α =  
  
  (ii) Decision:

  reject the null hypothesisdo not reject the null hypothesis     

  
  (iii) Reason for decision:

  Since α > p-value, we reject the null hypothesis.Since α < p-value, we do not reject the null hypothesis.     Since α > p-value, we do not reject the null hypothesis.Since α < p-value, we reject the null hypothesis.

  
  (iv) Conclusion:

  There is sufficient evidence to conclude that mean age at which American mothers wean their children is less than 4.There is not sufficient evidence to conclude that mean age at which American mothers wean their children is less than 4.     

• Part (i)

  Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Enter your answers in years. Round your answers to four decimal places.)

Answer 1

Ho :   µ =   4                  
Ha :   µ <   4       (Left tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s =    0                  
Sample Size ,   n =    21                  
Sample Mean,    x̅ =   1                  
                          
degree of freedom=   DF=n-1=   20                  
                          
Standard Error , SE = s/√n =   0.2500   / √    21   =   0.0546      
t-test statistic= (x̅ - µ )/SE = (   0.750   -   4   ) /    0.0546   =   -59.57
                             
                          
p-Value   =   0.0000   [Excel formula =t.dist(t-stat,df) ]              
Decision:   p-value<α, Reject null hypothesis                       

(iv) Conclusion:

There is sufficient evidence to conclude that mean age at which American mothers wean their children is less than 4.

.........................

THANKS

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