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[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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