Question

An analyst from an energy research institute in California wishes to precisely estimate a 99% confidence interval of the average price of unleaded gasoline in the state. In particular, she does not want the sample mean to deviate from the population mean by more than $0.06. What is the minimum number of gas stations that she should include in her sample if she uses the standard deviation estimate of $0.32, as reported in the popular press?

Answer 1

Answer:

190

Step-by-step explanation:

Data provided in the question:

Confidence level = 99%

Therefore,

α = 1% = 0.01

[ from standard normal table ]

z-value for $z_{\frac{\alpha}{2}}= z_{\frac{0.01}{2}}=$ = 2.58

Margin of error, E = $0.06

Standard deviation, σ = $0.32

Now,

n = $(\frac{z_{0.005}\sigma}{E})^2$

Here,

n is the sample size (or the minimum number of gas stations  )

on substituting the respective values, we get

= $(\frac{z_{0.005}\sigma}{E})^2$

= $(\frac{2.58\times0.32}{0.06})^2$

= 13.76²

= 189.3376 ≈ 190

Hence,

minimum number of gas stations that she should include in her sample is 190