Test Statistic Z-Value. Please help!

Question

100 people in New York and 100 people from Chicago, were each asked "do you like your own cities style of pizza better than the other one?"

The number of people that said yes in new york is (23+15+12) 50 people.

The number of people that said yes in Chicago was (20+5+18) 43 people.

The claim is made "opinions are different between the two cities."

1) Find this two-study comparison's test statistic z-value.

2)Using either a critical value or p-value comparison, what would you say about the claim?

Answer 1

You're trying to compare proportions from two independent samples: people from New York and people from Chicago, each answering whether they prefer their own city's pizza. The numbers in parentheses, like (23 + 15 + 12) = 50 and (20 + 5 + 18) = 43, are just different categories or subgroups that are being summed up. They might represent age groups, boroughs, etc., but you can ignore their individual meaning here — what matters is the total number of "yes" responses in each group:

1. New York: 50 said yes out of 100.
2. Chicago: 43 said yes out of 100.

So you're comparing:

1. p1 = 50/100 = 0.50 (New York)
2. p2 = 43/100 = 0.43 (Chicago)

Find the test statistic Z-Value:

We're doing a two-proportion z-test with null hypothesis.

H₀ : p1 = p2

H_a : p1 does not equal p2 (claiming opinions are different).

Compute the pooled proportion:

P^hat = (x₁ + x₂)/(n₁ + n₂) = (50+43)/(100+100) = 93/200 = 0.465

Compute the standard error (SE):

SE = √ p^hat(1 - p^hat)(1/n₁ + 1/n₂) = √(0.465)(0.535)(1/100 + 1/100) = √(0.248775)(0.02) = √0.0049755 = 0.0705

Compute z-value:

z = (p₁^hat - p₂^hat)/SE = (0.50 - 0.43)/(0.0705) = 0.07/0.0705 ≈ 0.993

What about the claim? Let's now assess whether the difference is statistically significant.

Option A: Use critical values

For a two-tailed test at significance level α = 0.05, the critical z-values are: ± 1.96.

Since our z-value is 0.993, which lies within the range -1.96 < z < 1.96, we fail to reject the null hypothesis.

Option B: Use p-value

A z-score of 0.993 corresponds to a p-value = 0.32 (from z-tables).

Since p = 0.32 > 0.05, again, we fail to reject the null hypothesis.

Final Conclusion:

There is not enough evidence to say that the opinions are different between the two cities. The observed difference could easily be due to random chance.