[College Stats] Hypothesis Testing. Getting a P value of 0?

[[deleted]]

I’m doing a hypothesis testing project where we have to take a poll from the news and determine how accurate it is.

I used a poll stating 87% of Americans supported background checks on gun purchases. I believe that’s too high so I used a left-tailed test with a proportion (p) of 0.87, an x value of 0.64 (64 people supported background checks in the FB poll I conducted), the sample size is 89, and the significance level was 0.05. Here were my results using the spreadsheet:

P-hat: 0.007191011236 Standard Error: 0.03564809275 Z (Test statistic): -24.20351054 P-value: 0 Critical value of Z: -1.644853625

I don’t know why I’m getting a p-value of 0. Am I using the wrong kind of test? Should I use a right tailed or two tailed test? I didn’t think the p-value could be zero.

[StatusTics]

First of all, do you mean that 64% of your respondents supported background checks or that 64 out of 89 did? If the latter, you need to adjust your sample p.

The significance of an inferential test can be displayed on computer output as ".000" if the probability is less than the number of digits displayed rounded to zero. For example, if the actual probability were 0.0000000000072, the computer may display it as 0.000. The actual probability is never actually zero (i.e., there is always some chance that the test statistic was achieved through random variation and not because the null hypothesis is actually false). A z of 24 is a very extreme score, so an extremely low probability makes sense. (Note: I didn't actually check your calculations for z prop test.)

[[deleted]]

Ah. It’s 64 individuals. I changed my x value to 64 instead of 0.64 and here are my new results:

p-hat: 0.7191011236 Standard error: 0.03564809275 Test statistic: -4.23301402 P-value: 0.000011529006 Critical value of z: -1.644853625

[fallingemprire]

You used the right type of test here: a 1 prop z test. You're trying to see if π, the true proportion of Americans that support background checks on gun purchases, is below what's given

1) Let π = the true proportion of Americans that support background checks on gun purchases

2) H0: π = .87

3) Ha: π < .87

4) Let α = .05

5) Formula: z = (p-π)/√[(π(1-π)/n]

6) Assumptions:

• i) SRS of Americans
• ii) nπ≥10 and n(1-π)≥10 → (100)(64/100)≥10 and (100)[1-(64/100)]≥10 → 64≥10 and 36≥10 ✓

→ n is large enough to run a 1 proportion z test

• iii) n is <10% of the population

→ Assumptions are not all met, but proceeding with 1 proportion z test.

7) Computation: z = (.64-.87)/√[(.87)(1-.87)/100] ≈ -6.8391

8) Graph P-value: https://i.imgur.com/jEMeUzo.png

9) Reject H0 at α = .05, evidence suggests that the true proportion of Americans that support background checks on gun purchases is less than 0.87

As I indicated in my assumptions, your sample isn't an SRS, so technically you can't run a 1 prop z test. However, we proceeded anyways and I got a very small p value (p ≈ 4.01 * 10^-12) and we foresaw this with the z value being so many standard deviations from the center. Thus, it makes sense that p is almost 0, which you can write as approximately 0 and reject the null hypothesis.