What is the test statistic, P value, and conclusion?

Question

In​ 1997, a survey of

880880

households showed that

134134

of them use​ e-mail. Use those sample results to test the claim that more than​ 15% of households use​ e-mail. Use a 0.05 significance level.

Answer 1

We have a claim about a population proportion, so this is a z-test.

Since the claim is that more than 15% of the population use email, our null and alternative hypotheses are:

H(null): p = 0.15

H(Alt): p > 0.15 (a right-tailed test)

We will use the results of the survey to calculate our point estimate, p-hat.

Sample size: n=880,880

Number of 'successes': x=134,134

p-hat = x/n = 134,134/880,880 = 0.1523

This value seems pretty close to 15%. Do you want to take a wild guess as to whether we will accept or reject our null hypothesis? Maybe you'd better wait until we are done.

Before we can calculate our test statistic, we need to calculate the standard error.

Our standard error is the square root of the theoretical probability of success (p) times the theoretical probability of failure (1-p) divided by the sample size. Please don't confuse p with p-hat. p is the value stated in the hypotheses.

SE=sqrt(0.15 x 0.85 / 880,880) = 0.00038

Now we can calculate the test statistic.

z = (p-hat - p) / standard error = (0.1523- 0.15 ) / .00038 = 6.05

Wow! That is unusual z-score. Literally off the charts. The p-value is the probability of getting a z-score more extreme than 6.05. What is that probability? Most z-tables only show z-scores between -3.5 and 3.5. For z-scores greater than 3.5 or less than -3.5, we can state that the p-value is < 0.0001.

Certainly 0.0001 < 0.05 (the critical value or alpha), so we will reject the null hypothesis. The evidence supports the alternative hypothesis, the claim that more than 15% of households use email.

Does this result surprise you? Recall our point estimate, p-hat was pretty close to 15%. But think about the sample size. It is HUGE! Therefor our standard error is super small, resulting in a large z-score.

Hope that helps!

Answer 2

Ho: p = .15, Ha: p > 0.15, x = 134134, n = 880880, p_hat = x/n = 0.1522727, alpha = .05.

We have a very large number of households in our sample and the sample p_hat is very close to the hypothesized value of 0.15.

This proportion will be normally distributed if we kept taking these similarly sized samples again and again with standard error, SE, of sqrt(p*(1-p)/n) = sqrt(.15*.85/ 880880) = 0.0003804492 = SE.

Now we can calculate a z score for our particular p_hat:

z = (p_hat - p) / SE = 5.973799. So it turns out that 15.22 is really very unlikely given p is .15 and the number in the survey n is so huge. Almost all the variation will go away with a sample that large!

No need to go to a z table because z was so huge. We will reject Ho at the 5% level and conclude that there is evidence to support the claim that the true proportion is bigger than p = .15. Note that this was technically a one sided test. If the z score had been less intuitively extreme, we might have had to actually do work and calculate a p-value to compare to alpha. But if you have to answer the question about p-value and not just make a conclusion, you can say that the p-value is ~0

(it's actually 0.00000000158957.)

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