Long Division: Write down your steps! 7_984
1 answer:
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Answer:
95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].
Step-by-step explanation:
We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.
Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of residents 65-years or older who had obtained a flu shot =
= 0.25
n = sample of residents 65-years or older = 111
p = population proportion of residents who were getting the flu shot
<em>Here for constructing 95% confidence interval we have used One-sample z test for proportions.</em>
<u>So, 95% confidence interval for the population proportion, p is ;</u>
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.169 , 0.331]
Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].