Suppose a survey research firm selected a simple random sample of 300 people for the purposes of estimating their mean weekly income and found that the sample mean was $1,345.78 with a standard deviation of $257.90
Calculate the 90% confidence interval for the population mean.
In this case the sampling distribution should be approximately normally distributed. Because the sample size (300) is large, we know from the central limit theorem that the sampling distribution of the mean will be normal or nearly normal.
Now we can use the following steps to construct a 90% confidence interval.
t0.05(299) = 1.9679 (two tail)
ME = critical value * standard error
= 14.8899 * 1.9679
= 29.3018
Now, confidence interval is defined by the sample statistic + margin of error
So, 95% confidence interval for population mean :
1345.78 + 29.3018
CI = ( 1316.4782 , 1375.0818)
Suppose a survey research firm selected a simple random sample of 300 people for the purposes...
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