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Answer:
95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].
Step-by-step explanation:
We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.
A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.
Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of males having blood disorder=
= 0.25
= sample proportion of females having blood disorder =
= 0.275
= sample of males = 1000
= sample of females = 1000
= population proportion of males having blood disorder
= population proportion of females having blood disorder
<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>
<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u><u>)</u><u> 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( < (
) <
) = 0.95
<u>95% confidence interval for</u> () =
[,
]
= [ ,
]
= [-0.064 , 0.014]
Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].