Question

A sample of 2500 gamers finds that 48% are women. Determine whether half of gamers are women at the 5% level. How about the 1% level?

a. Be sure to carefully identify the null and alternative hypotheses.
b. What are the test statistic and critical value?
c. What is the result of the test (with respect to the null hypothesis). Include a statement of the null and/or alternative when appropriate.
d. Construct C.I at the same confidence level and interpret it
e. How would (d) be useful in (a)-(c) above?

Answer 1

Part (a)

p = population proportion of gamers that are women

Null: p = 0.50

Alternate: p ≠ 0.50

This is a two-tailed test because of the "not equals" in the alternate hypothesis.

==========================================

Part (b)

SE = Standard Error

SE = sqrt(p*(1-p)/n)

SE = sqrt(0.5*(1-0.5)/2500)

SE = 0.01

Test statistic:

z = (phat - p)/SE

z = (0.48 - 0.50)/0.01

z = -2

At the 5% significance level, the critical z values for a two-tailed test are: -1.96 and 1.96. These values are found in a Z table. Or use a stats calculator.

At the 1% significance level, the critical z values for a two-tailed test are: -2.576 and 2.576

==========================================

Part (c)

At the 5% significance level, we found the critical z values -1.96 and 1.96

The test statistic (z = -2) is NOT between those critical values, so this value is in the rejection region. We reject the null and conclude that the alternate hypothesis is the case. We conclude that p ≠ 0.50; either p < 0.50 or p > 0.50

Based on phat = 0.48, it appears that p < 0.50 might be the case.

-----------

Now switch to the 1% significance level. The critical z values are roughly -2.576 and 2.576

We see that the test statistic (z = -2) is between those critical values. This time we fail to reject the null. The conclusion at the 1% significance level is "it appears 50% of the gamers are women".

As you can see, adjusting the significance level sometimes will adjust the conclusion to what the researchers want/expect to see.

==========================================

Part (d)

alpha = significance level = 0.05

C = confidence level

C = 1-alpha

C = 1 - 0.05

C = 0.95

A significance level of 5% leads to a 95% confidence level.

E = margin of error

E = z*sqrt(phat*(1-phat)/n)

E = 1.96*sqrt(0.48*(1-0.48)/2500)

E = 0.019584 approximately

L = lower boundary of confidence interval

L = phat - E

L = 0.48 - 0.019584

L = 0.460416

L = 0.4604

U = lower boundary of confidence interval

U = phat + E

U = 0.48 + 0.019584

U = 0.499584

U = 0.4996

The 95% confidence interval is roughly (L,U) = (0.4604,0.4996)

p = 0.50 is not between those endpoints, so we reject the null.

-------------------------

Recalculate the confidence interval boundaries, but this time at the 1% significance level (aka 99% confidence). I'll skip the steps.

You should get roughly (0.4543, 0.5057)

This time p = 0.50 is in the interval, so we fail to reject the null.

==========================================

Part (e)

Part (d) is useful to see another viewpoint why we either reject the null or fail to reject the null. This avoids having to compute the test statistic. The drawback is that you do a bit more calculations, and you still need the critical values.