Answer:
The 99% C.I is ( 367.703, 382.297
)
Step-by-step explanation:
Given that:
sample mean [tex]\overline x[/tex] = 375
Population standard deviation [tex]\sigma[/tex] = 20
Sample size n = 50
The level of significance [tex]\alpha[/tex] = 1 - C.I = 1 - 0.99 = 0.01
From z tables:
The critical value = [tex]Z_{\alpha /2}[/tex] = [tex]Z_{0.005}[/tex] = 2.58
The margin of error E = [tex]Z_{\alpha /2}[/tex] [tex]\times \dfrac{\sigma }{\sqrt{n}}[/tex]
The margin of error E = [tex]2.58 \times \dfrac{20}{\sqrt{50}}[/tex]
The margin of error E = 7.297
The limits of 99% C.I are expressed as:
Lower limit = [tex]\overline x - E[/tex] = 375 - 7.297
Lower limit = 367.703
Upper limit = [tex]\overline x + E[/tex] = 375 + 7.297
Upper limit = 382.297
The 99% C.I is [tex]\overline x \pm E[/tex] = [tex]375 \pm 7.297[/tex]
= ( 367.703, 382.297
)
