Answer:
(21.4, 18.6)
Step-by-step explanation:
Given that:
Mean (μ) = $20 per week
Standard deviation (σ) = $5
number of sample (n) = 49
Confidence level = 95% = 0.95
α = 1 - 0.95 = 0.05
[tex]\frac{\alpha }{2}=\frac{0.05}{2} = 0.025\\ z_{\frac{\alpha }{2} }=z_{0.025}= 1.96[/tex] from the probability table.
The margin of error (e) = [tex]z_{\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } = 1.96*\frac{5}{\sqrt{49} } =1.4[/tex]
Confidence interval = μ ± e = (μ + e, μ - e) = (20 + 1.4, 20 - 1.4) = (21.4, 18.6)
