A news article estimated that only 5% of those age 65 and older who prefer to watch the news, rather than to read or
listen, watch the news online. This estimate was based on a survey of a large sample of adult Americans. Consider
the population consisting of all adult Americans age 65 and older who prefer to watch the news, and suppose that for
this population the actual proportion who prefer to watch online is 0.05.
A random sample of n = 100 people will be selected from this population and p̂, the proportion of people who
prefer to watch online, will be calculated. What are the mean and standard deviation of the sampling
distribution of p̂? (Round your standard deviation to four decimal places.)
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The mean of the sampling distribution of p̂ is equal to the actual proportion of people who prefer to watch online, which is 0.05.
The standard deviation of the sampling distribution of p̂ can be calculated using the formula:
Standard deviation = sqrt(p*(1-p)/n)
where p is the actual proportion of people who prefer to watch online, and n is the sample size.
In this case, p = 0.05 and n = 100.
Standard deviation = sqrt(0.05*(1-0.05)/100)
= sqrt(0.0475/100)
= sqrt(0.000475)
≈ 0.0218
Therefore, the standard deviation of the sampling distribution of p̂ is approximately 0.0218.
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