a. 0.95
b. 99%
c. 50.0
d. 90 - 95
2.What is the effect of the level of confidence on the confidence interval?
3.What is the margin of error E?
4.Given the information that a sampled population is normally distributed with X = 36, σ = 3 and n = 20.
a. What is the 95% confidence interval for μ?
b. Are the assumptions met? Explain
1. b. 99%
2. Increasing the confidence will increase the margin of error resulting in a wider interval. Increasing the confidence will decrease the margin of error resulting in a narrower interval.
3. The margin of error is a statistic expressing the amount of random sampling error in the results of a survey. The larger the margin of error, the less confidence one should have that a poll result would reflect the result of a survey of the entire population.
4. n = 20
"\\bar{X} = 36"
σ = 3
In this case, the level of confidence can be defined as,
1 - α = 0.95
α = 0.05
The value of the confidence coefficient with 95% level of confidence is computed by using the standard normal z table,
"z_{\u03b1\/2} = z_{0.05\/2} \\\\\n\n= z_{0.025} \\\\\n\n= 1.96"
a. The 95% confidence interval for the mean is
"36 - (1.96)\\frac{3}{\\sqrt{20}} < \u03bc < 36 + (1.96) \\frac{3}{\\sqrt{20}} \\\\\n\n36 - 1.31 < \u03bc < 36 + 1.31 \\\\\n\n34.69 < \u03bc < 37.31"
b. There are 2 requirements to use the x-proportion interval: random sample and large enough sample.
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