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A survey conducted by a reputed Business School in Texas among 400 students showed that students spend on an average $500 for textbooks in a year. The sample standard deviation was $80. What is the confidence interval for the average amount spent on textbooks in a year if the desired level of confidence is 99%

Sagot :

[489.68, 510.32] is the confidence interval for the average amount spent on textbooks in a year if the desired level of confidence is 99%.

n=400

sample mean (x)=500

sample standard deviation (s)=80

99% confidence interval for mean

α=0.01

critical value [tex]Z_{\frac{a}{z} }[/tex] =2.58

Standard error of sample mean Sx = [tex]\frac{s}{\sqrt[]{n} }[/tex] = [tex]\frac{80}{\sqrt[]{400} }[/tex] = [tex]\frac{80}{20}[/tex] = [tex]4[/tex]

∴ Sx = 4

Confidence interval (μ)

μ= [tex]x[/tex] ± [tex]Z_{\frac{a}{z} }[/tex] [tex]\frac{s}{\sqrt[]{n} }\\[/tex]

μ= [tex]x[/tex] ± [tex]Z_{\frac{a}{z} }[/tex] (Sx)

=500 ± 2.58(4)

=500 ±  10.32

lower limit = 500 - 10.32 = 489.68

upper limit= 500 + 10.32= 510.32

Hence, [489.68, 510.32] is the confidence interval for the average amount spent on textbooks in a year if the desired level of confidence is 99%.

Learn more about confidence interval here https://brainly.com/question/15712887

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