Answer:
Answer = √(0.301 × 0.699 / 83) ≈ 0.050
A 68 percent confidence interval for the proportion of persons who work less than 40 hours per week is (0.251, 0.351), or equivalently (25.1%, 35.1%)
Step-by-step explanation:
√(0.301 × 0.699 / 83) ≈ 0.050
We have a large sample size of n = 83 respondents. Let p be the true proportion of persons who work less than 40 hours per week. A point estimate of p is because about 30.1 percent of the sample work less than 40 hours per week. We can estimate the standard deviation of as . A confidence interval is given by , then, a 68% confidence interval is , i.e., , i.e., (0.251, 0.351). is the value that satisfies that there is an area of 0.16 above this and under the standard normal curve.The standard error for a proportion is √(pq/n), where q=1−p.
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Mark brainliest
Is this a true/false? if so, it's true. (Only SAS, SSS, ASA, and AAS prove congruency between triangles).
Hope it helps <3
Answer:
1.5707
Step-by-step explanation:
i solved it using the formula :
Answer:
<em>31/50 62%</em>
<em>0.6 30/50 60%</em>
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Hope it works out for you!
