A recent survey of 50 executives who were laid off during a recent recession revealed it took a mean of 26 weeks for them to find another position. The standard deviation of the sample was 6.2 weeks. Construct a 95% confidence interval for the population mean.

Respuesta :

Answer: [tex](24.28,\ 27.72)[/tex]

Step-by-step explanation:

Given : Sample size : [tex]n=50[/tex]

Sample mean : [tex]\overline{x}=26[/tex]

Standard deviation : [tex]\sigma =6.2[/tex]

Significance level : [tex]\alpha=1-0.95=0.05[/tex]

Critical value : [tex]z_{\alpha/2}=1.96[/tex]

Formula to find the confidence interval for population mean :-

[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\\\\=26\pm(1.96)\dfrac{6.2}{\sqrt{50}}\\\\\approx26\pm1.72\\\\=(26-1.72,\ 26+1.72)\\\\=(24.28,\ 27.72)[/tex]

Hence, a 95% confidence interval for the population mean = [tex](24.28,\ 27.72)[/tex]

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