Question

i meed help with b-f please

1) Public health officials believe that 98% of children have been vaccinated against measles. A random survey of medical records at many schools across the country found that, among more than 13,000 children only 97.4% had been vaccinated. Has the percentage of children vaccinated against measles decreased? Use a significance level of a = 0.05. a) Write the hypotheses. (2 points) b) Check the necessary assumptions and conditions. (4 points) c) Perform the appropriate hypothesis test using a significance level of a = 0.05, report the test statistic and p-value, and write a full conclusion. Be sure to write out what you entered in your calculator. (5 points) d) Are the results statistically significant? Why or why not? (2 points) e) Find the appropriate 90% confidence interval. Be sure to write out what you entered in your calculator. (3 points) f) Are the results practically significant? Why or why not? (2 points)

Answer 1

a)

Ho :   p =    0.98  
H1 :   p <   0.98   (Left tail test)

b)

sample size,n < 5% of population size,N
np>5 and n(1-p)>5

c)
          
Level of Significance,   α =    0.05  
Number of Items of Interest,   x =   12662  
Sample Size,   n =    13000  
          
Sample Proportion ,    p̂ = x/n =    0.9740  
          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0012  
Z Test Statistic = ( p̂-p)/SE =    (0.974-0.98)/0.0012=   -4.89  
          
          
p-Value   =   0.0000   [excel function =NORMSDIST(z)]
Decision:   p-value<α , reject null hypothesis       
There is enough evidence that proportion of measeles decreased

d)
Significant because we reject the hypothesis

e)

Level of Significance,   α =    0.10          
Number of Items of Interest,   x =   12662          
Sample Size,   n =    13000          
                  
Sample Proportion ,    p̂ = x/n =    0.9740          
z -value =   Zα/2 =    1.645   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.001396          
margin of error , E = Z*SE =    1.645   *   0.00140   =   0.0023
                  
90%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.97400   -   0.00230   =   0.9717
Interval Upper Limit = p̂ + E =   0.97400   +   0.00230   =   0.9763
                  
90%   confidence interval is (   0.972   < p <    0.976   )

f)

Results are not practically significant because the change is very minor

Please let me know in case of any doubt.

Thanks in advance!

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