Hypothesis Tests – Calculation of P-Values
Example 1: Directional hypothesis test for a population mean.
The U. S. Food and Drug Administration recommends that individuals consume 1000 mg of
calcium daily. The International Dairy Foods Association (IDFA) sponsors an advertising
campaign aimed at male teenagers to try to increase calcium consumption. After the campaign, the
IDFA obtained a random sample of 50 male teenagers and found that the mean amount of calcium
consumed was x  1081 mg, with a standard deviation of s = 426 mg. Conduct a test to determine
whether the campaign was effective. Use the  = 0.05 level of significance.
We want to prove that the mean, , calcium consumption for male teenagers is greater than
the FDA recommendation of 1000 mg per day. Hence,
Step 1: H0:   1000 mg. versus
Ha:  > 1000 mg.
Note that the alternative hypothesis states what we are trying to prove; the null hypothesis
states the opposite.
Step 2: n = 50,  = 0.05
Step 3: Since we are testing a hypothesis about a population mean, the test statistic is
X  1000 mg
Z
, which under H0 has an approximate standard normal distribution
 S 


 50 
Step 4: The alternative hypothesis says “greater than.” Therefore the rejection region is a righthand tail of the standard normal distribution. The area of this right-hand tail is 0.05, our chosen
significance level. The rejection region looks like:
Step 5: Now, we select the random sample from the population, collect the data, and do the
calculations. We find that x  1081 mg, s = 426 mg, z = 1.3445.
We want to calculate the p-value for this result. We have
𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 𝑃(𝑍 ≥ 1.3445) = 0.0894.
The p-value is greater than our chosen significance level.
Step 6: We fail to reject H0 at the 0.05 level of significance. We do not have sufficient evidence to
conclude that the mean daily calcium consumption by male teenagers is greater than the FDA
recommended amount of 1000 mg.
Example 2: Hypothesis test about a population proportion.
In 1995, 40% of adults aged 18 years or older reported that they had “a great deal” of
confidence in the public schools. On June 1, 2005, the Gallup Organization released results of a
poll in which 372 of 1004 adults aged 18 years or older stated that they had “a great deal” of
confidence in the public schools. Does the evidence suggest that the proportion of adults aged 18
years or older having “a great deal” of confidence in the public schools has decreased between 1995
and 2005? Use  = 0.05.
Since we are testing hypotheses about a population proportion, we are collecting the data
using a binomial experiment. There are 1004 trials in the experiment. The trials are independent of
each other due to random sampling. The trials are identical to each other because each trial consists
of randomly selecting an adult and asking whether he/she has “a great deal” of confidence in the
public schools. There are two possible outcomes for each trial: either the person says, “Yes” or the
person says, “No.” Due to random sampling, the probability of success is the same for each trial,
namely the fraction of the population who would say, “Yes.”
Step 1: H0: p  0.40
Step 2: n = 1004,  = 0.05
versus
Step 3: The test statistic is Z 
pˆ  0.40
Ha: p < 0.40
0.400.60
, which under H0 has an approximate standard
1004
normal distribution.
Step 4: The rejection region is:a left-hand tail of the standard normal distribution, with area 0.05.
Step 5: Now, we select the random sample from the population, collect the data, and do the
372
 0.3705 , z = -1.9069.
calculations. We find that pˆ 
1004
We want to calculate the p-value for this result. We have
𝑝 − 𝑣𝑎𝑙𝑢𝑒 = 𝑃(𝑍 ≤ −1.9069) = 0.0283.
The p-value is less than our chosen significance level.
Step 6: We reject H0 at the 0.05 level of significance. We have sufficient evidence to conclude that
the fraction of adults 18 years or older who have “a great deal” of confidence in the public schools
decreased between 1995 and 2005.
In this last example, note that if we had chosen α = 0.01, then the data would have led us to fail to
reject the null hypothesis.