Question

# In a recent random sample, where 600 adults were surveyed, 16.4% indicate that they have fallen...

In a recent random sample, where 600 adults were surveyed, 16.4% indicate that they have fallen asleep watching television programs at night. Find the 98% confidence interval for the population proportion of people who sleep watching television at night.

Solution:
Given in the question
No. of adults surveyed = 600
P(they have fallen asleep watching television programs at night) = 0.164
P(they did not fall asleep watching television programs at night) = 1 - 0.164 = 0.836
Here we will use one proportion Z test because np= 98.4 and np(1-p) =82.3 is greater than 10
We need to calculate 98% confidence interval so alpha= 0.02
alpha/2 = 0.01
Zalpha/2 from Z table is 2.3264
98% confidence interval can be calculated as
p +/- Zalpha/2*Sqrt(p*(1-p)/n)
0.164 +/- 2.3264*sqrt(0.164*(1-0.164)/600)
0.164 +/- 2.3264*0.015
0.164 +/- 0.035
So 98% confidence interval is 0.129 to 0.199
Therefore based on data provided 98% confidence interval for the population proportion of people who sleep watching television at night is 0.129 to 0.199.

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