A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard deviation of 34. What is the 90% confidence interval for the population mean? Use the table below to help you answer the question.
Answer:
Step-by-step explanation:
Given that A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard deviation of 34.
Confidence level = 90%
Z critical value = 1.645
Sample size n=90
Std error of sample = sigma/sqrt n= 34/9.487
=3.583
Margin of error = ±1.645(3.583)=5.895
Hence confidence interval lower bound = 138-5.895 = 132.105
Upper bound = 138+5.895 = 143.895
Hence confidence interval = (132.11, 143.90)
We can be 90% confident that the population mean (μ) falls between 132.1 and 143.9.
Given to us
The standard deviation tells us about the variation of the data point from the mean.
\(S_M = \sqrt{\dfrac{\sigma^2}{90}}\\\\S_M = \sqrt{\dfrac{34^2}{90}}\\\\S_M = 3.58\)
The 90% confidence interval for the population mean,
\(\mu = M \pm Z(S_M)\)
Substitute the values,
\(\mu = M \pm Z(S_M)\\\\\mu = 138 \pm (1.64\times3.58)\\\\\mu = 138 \pm 5.9\\\\\)
\(\mu = 138, 90\%\ CI\ [132.1, 143.9].\)
Hence, we can be 90% confident that the population mean (μ) falls between 132.1 and 143.9.
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