Answer :
lower limit of the 99% interval = 3.7265
Upper limit of the 99% interval = 6.1535
What do you mean by standard deviation?
The standard deviation is a quantity that reveals the degree of variation (such as spread, dispersion, and spread) from the mean.
Number of students, n = 11
Averange time , mean , μ = 4.94
Standard deviation , [tex]s_{x}[/tex] = 1.27
Significance level , α = 1 -= 0.99 = 0.01
Degree of freedom for t-distribution , d.f. = n - 1 = 11 - 1 = 10
Critical value = [tex]t_{\alpha /2,d.f[/tex] = [tex]t_{0.005,10}[/tex] = 3.169
Margin of error ,E = [tex]t_{\alpha /2,d.f[/tex] × [tex]\frac{s_{x} }{\sqrt{n} }[/tex] = 3.169 ×[tex]\frac{1.27}{\sqrt{11} }[/tex] = 3.169 × 0.38292 = 1.21347
Limits of 99% confidence intervals are given by:
a) Lower limit = μ - E = 4.94 - 1.21347 = 3.7265
b) Upper limit = μ + E = 4.94 + 1.21347 = 6.1535
Therefore, lower limit of the 99% interval = 3.7265
Upper limit of the 99% interval = 6.1535
To learn more about the stanadrd deviation from the given link.
https://brainly.com/question/475676
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