At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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
#SPJ4
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
