Find the 95% confidence interval and explain. O 0.40 - (0.0223)(1.96) = 0.3563 and 0.40 + (0.0223)(1.96) = 0.4437. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 35.63% and 44.37%. O 0.453 - (0.0219)(1.645) = 0.4170 and 0.453 + (0.0219)(1.645) = 0.4890. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 41.7% and 48.9%. O 0.453 - (0.0223)(1.96) = 0.4093 and 0.453 + (0.0223)(1.96) = 0.4967. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 40.93% and 49.67% O 0.40 - (0.0223)(1.645) = 0.3633 and 0.40 + (0.0223)(1.645) = 0.4367. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 36.33% and 43.67%.
According to the Center for Disease Control (CDC) during the 2018-2019 season the percent of Americans 18 and older who had a flu shot was 45.3%. We did a random sample of 500 Americans and found that 200 out of the 500 had a flu shot. Is there enough evidence at the 5% level that 45.3% or less of Americans 18 or older had a flu shot?
Find the 95% confidence interval and explain.
a) 0.40 – (0.0223)(1.96) = 0.3563 and 0.40 + (0.0223)(1.96) = 0.4437. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 35.63% and 44.37%.
b) 0.453 – (0.0219)(1.645) = 0.4170 and 0.453 + (0.0219)(1.645) = 0.4890. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 41.7% and 48.9%.
c) 0.453 – (0.0223)(1.96) = 0.4093 and 0.453 + (0.0223)(1.96) = 0.4967. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 40.93% and 49.67%
d) 0.40 – (0.0223)(1.645) = 0.3633 and 0.40 + (0.0223)(1.645) = 0.4367. We are 95% confident the proportion of Americans 18 and older who had a flu shot is between 36.33% and 43.67%.
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