A random sample of 1236 Internet users wasasked where they will go for informati

Question

     A random sample of 1236 Internet users wasasked where they will go for information the next time they needinformation about health or medicine; 768 said that they would usethe Internet. Calculate and interpret a 90 % confidence intervalfor the proportion of all Internet users who would use the Internetnext time for information about health or medicine.

(a) Describe the population of interest and explain in words whatthe parameter p is.

(b) Give the numerical value of the sample proportion of successesthat estimates p.

(c) Calculate the 90% confidence interval for p.

(d) Interpret the interval.

Explanation / Answer

(a) The population of interest is all Internet users.

      The parameter p is the proportion ofall internet users who will go for information the next time theyneed information about health or medicine. (b) p=768/1236 =0.62 (c) Given =0.1, |Z(0.05)|=1.645 (check normal table)

The 90% CI is
p ± Z*p*(1-p)/n
--> 0.62 ± 1.645*sqrt(0.62*(1-0.62)/1236)
--> (0.597, 0.643)
(d) Based on (c), we have 90% confidence that the truepopluation p will be in this interval.

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