Null hypothesis: $p=60 $% Alternative hypothesis: $p>60$%

$5$% level of significance
proportion = $\frac{382}{500} = 0.764$

The equation I used to find my Z-test statistic is
$$z = \dfrac{\text{proportion} - p}{ \text{standard deviation}}$$

so : $\text{standard deviation} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.6(1-0.6)}{500}} = 0.0219$

then $z= \frac{0.764-0.6}{0.0219} = 7.49$

I know there is something wrong with my Z-test statistic because I need to use my Z-test statistic to find the $p$-value. Would someone please let me know what I am doing wrong?

  • $\begingroup$ I really don't see what's wrong with your method above. Your $p$-value is going to be very small. Nothing wrong with that. $\endgroup$ – Clarinetist Sep 16 '15 at 5:37

As @Clarinetist says, you have done this correctly. Here is some additional information.

Your observed 76.4% seems much larger than the hypothetical 60%.

Reality check: A very rough rule of thumb for a 2-sided confidence interval is the observed percentage $\pm 1/\sqrt{n}$. For $n = 500,$ that's about $\pm 4.5\%,$ so a true value of 60% seems far from believable. This rough rule is often used to give margins of sampling error for public opinion polls. (Maybe you've seen $\pm 3\%$ as the margin of error for a poll with about 1100 subjects.)

Your P-value will be far beyond the values in a printed normal table and so essentially 0 (strong evidence for rejecting $H_0; p = 0.6$ against $H_a: p > 0.6$). In R statistical software, the P-value of your one-sided test using the normal approximation can be found as shown below. The last value given (based on your computations) is harmlessly a little different because of rounding error:

 std.err = sqrt(.6*.4/500);  std.err
 ## 0.02190890
 1 - pnorm(382/500, .6, std.err)
 ## 3.563816e-14
 1 - pnorm(7.49)   # Your standardized version
 ## 3.441691e-14

An exact P-value can be obtained from the binomial distribution (without using the normal approximation).

 binom.test(382, 500, .6, alternative="greater")

         Exact binomial test

 data:  382 and 500 
 number of successes = 382, number of trials = 500, 
    p-value = 6.928e-15
 alternative hypothesis: true probability of success is greater than 0.6 
 95 percent confidence interval:
  0.7306433 1.0000000 
 sample estimates:
 probability of success 

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