Question

According to a report issued by the CDC in 2013, 66.8 % of women 40 years of age and over had a mammogram within the last two years. An oncologist at the CDC would like to know if there has been a change in this percentage over the last two years. A random sample of 648 women (40 and older) is taken, and each is asked the date of their last mammogram.

• Verify that the assumptions have been satisfied to conduct a hypothesis test.
  • Random sample?  (Yes, No)
  • The variable recorded was: (Mammogram occurred in last two years, Percentage of women who have had a mammogram in the last two years, Number of mammograms received, Date of last mammogram)
  • np =  and n(1 - p) = . Both of the values are greater than 10: (False, True)

It was found that 423 women in the sample had a mammogram in the last two years. Test the appropriate hypotheses using a significance level of 0.05.

• H0:H0:  (p, p̂) (>, <, ≠, = ) Ha:Ha:  (p̂, p) (> ≠ <)  
• αα =    decision rule: reject H0H0 if probability ? < > ≠   αα
• Test Statistic: z =   (Note: round the z-score to two decimal places - carry at least four decimal places throughout all of your calculations)
• probability =   (Note: round the probability to four decimal places)
• Decision: Reject H₀ OR Fail to reject H₀
• Conclusion: At the 0.05 level, there (is not, is) significant evidence to conclude the percentage of women 40 or older who have had a mammogram in the last two years is (less than, different than, greater than) 66.8%.

Answer 1

• Random sample = Yes
• The variable recorded was: (Mammogram occurred in last two years)
• np =    648   *   0.668   =           432.864
  n(1-p) =    648   *  ( 1 -    0.668   ) =   215.1360
  
• Both of the values are greater than 10: (True)

Ho :   p =    0.668                  
H1 :   p ╪   0.668       (Two tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   423                  
Sample Size,   n =    648                  
                          
Sample Proportion ,    p̂ = x/n =    0.6528                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0185                  
Z Test Statistic = ( p̂-p)/SE = (   0.6528   -   0.668   ) /   0.0185   =   -0.8228
                          

                          
p-Value   =   0.410606646   [excel formula =2*NORMSDIST(z)]              
Decision:   p value>α ,do not reject null hypothesis       

              
Conclusion: At the 0.05 level, there is not significant evidence to conclude the percentage of women 40 or older who have had a mammogram in the last two years is different than 66.8%.

Thanks in advance!

revert back for doubt

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