Hypothesis test: Find the critical value of Z, Zc

[so_gr_lo]

Homework Statement

Question:
The national price of a good is distributed normally with mean £1.20, standard deviation 5p. Sample data is for price in London, determine whether there is sufficient evidence at 5% level to suggest that the mean price in London is different to the national mean. Standard deviation same for both.

Relevant Equations

Test statistic = sample mean X - μ /sqrt(σ^2 / n) n = sample size

Question:
The national price of a good is distributed normally with mean £1.20, standard deviation 5p. Sample data is for price in London, determine whether there is sufficient evidence at 5% level to suggest that the mean price in London is different to the national mean. Standard deviation same for both.

attempt at solution:

H₀: μ = 1.20
H₁:μ≠ 1.20 (two -tailed test)

standard deviation: 0.05
national price mean: 1.20
sample data mean: μ = 1.176
Test statistic: Z = -1.518

How can I find the critical value of Z, Zc, on a calculator (CASIO fx-991EX classwiz). And how would I compare it to the test statistic to Zc in order to test the hypothesis. Do I test if -1.528 > Zc or -1.528 < Zc, not sure how to do it for two-tailed test.

thanks

 

[jambaugh]

Minor point of notation. You should use \overline{X} ("Xbar") for the sample mean to distinguish it from the population (distribution) mean.

The critical Z-value for a two tailed test will be the number of standard deviations Zc such that the probability that a standard normal RV is within that range is 1 - significance i.e the probability it is beyond that range is 5%.
P(-Zc < Z < Zc) = 95% i.e. P( Z<-Zc or Zc < Z) = 5%.

I am not familiar with your model but if it has an invNorm function that will give you invNorm(p)= z such
that P(Z < z) = p. To use this to get the two tailed range remember that the probability of being within is, in your case 95% and being beyond is 5%. Due to symmetry half that 5% is above and half is below the middle range. Thus P(Z < Zc) =(1 - 5%/2) (1 - upper tail probability).