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A screening program for neuroblastoma (a type of cancer) was undertaken in Germany among children born...

A screening program for neuroblastoma (a type of cancer) was undertaken in Germany among children born between November 1, 1993, and June 30, 2000, who were between 9 and 18 months of age between May 1995 and April 2000. A total of 1,475,773 children participated in the screening program. Of whom 204 were diagnosed between 12 and 60 months of age. The researchers expected the incidence rate of neuroblastoma to be 7.3 per 100,000 children during this period in the absence of screening. We wish to test if the number of cases detected by the screening program is significantly greater than expected.

a) Write hypotheses to test this claim. Explain why you should use a one sided alternative.

b) You may assume any necessary conditions have been met. Perform your test.

c) Do you think that the number of cases detected by the screening program is significantly greater than expected? Explain.

d) Give a 95% confidence interval for the incidence rate of neuroblastoma in the screened population.

e) Express your confidence interval from part d) as (p1, p2), where p1 and p2 are in the units of number of cases per 100,000 children.

specifically need help with question e

Solutions

Expert Solution

Solution

a.)

Because We wish to test if the number of cases detected by the screening program is significantly greater than expected.

b.)Samples drawn from the population is normally distributed and sample size is very large so we can use single proportion test.

c.)

=

p-value < significance(alpha=0.05 default), Reject the null hypothesis.

There is enough evidence that the screening program is significantly greater than the expected range at 0.05 significance level.

d.)

critical value at 95% confidence(Zcritical)=1.96

95% confidence Interval is given by:

Ans: (0.00002,0.000126)

e.)we can see in the above confidence interval, this interval does not include 'Zero' Hence null hypothesis should be rejected.

Hence

There is enough evidence that the screening program is significantly greater than the expected range at 0.05 significance level.


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