Question

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.

Answer 1

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.

• Identify a sample statistic: Since we are trying to estimate a population mean, we choose the sample mean (1345.78) as the sample statistic.
• Select a confidence level: In this context  the confidence level is 95% i.e. α = 0.05
• Find standard deviation or standard error : Since the standard deviation of the population is unknown, we cannot compute the standard deviation of the sample mean, instead, we compute the standard error (SE).

$SE =\frac{s}{\sqrt{n}}$

   $\Rightarrow SE =\frac{257.90}{\sqrt{300}}= 14.8899$

• Find critical value : The critical value is t score at α = 0.05 for d.f. = n - 1 = 300 = 299

t_0.05(299) = 1.9679 (two tail)

• Compute margin of error (ME) :

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 :
  
$\Rightarrow$ 1345.78 +  29.3018

CI = ( 1316.4782 , 1375.0818)