Relationship Between Hypothesis Tests and Confidence Intervals
Example 1: Non-directional hypothesis test for a population mean.
Suppose the quality control executive for the Post Company wants to test whether 24-oz.
packages of Post Grape-Nuts contain, on average, an amount of product differing from the target
value of 24 oz.. She selects a random sample of 100 boxes of cereal and measures the amount of
product in each one. The sample average is found to be 𝑦̅ = 23.94 oz., and the sample standard
deviation is found to be s = 0.13 oz. Test the claim at the 0.05 level of significance.
Step 1: H0:  = 24 oz.
versus
Ha:   24 oz.
Step 2: n = 100,  = 0.05
Note that in the real world, we would decide on a value for  based on our examination of
the possible consequences of the two possible types of mistakes we could make – a Type I error or a
Type II error. In this situation, a Type I error would be that we concluded that the box-filling
process is faulty, when in fact it is not. Possible consequences might be unnecessary down-time for
their assembly line. A Type II error would be to fail to conclude that the box-filling process is
faulty, when in fact it is. The Post company would not correct the problem, and consumers either
the company would continue to lose money or customers would be short-changed.
Step 3: Since we are testing a hypothesis about a population mean, the test statistic is
Y  24 oz
Z
, which under H0 has an approximate standard normal distribution.
 S 


 100 
Step 4: The alternative hypothesis says “not equal.” Therefore the rejection region is the union of a
left-hand tail of the standard normal distribution and a right-hand tail of the distribution, each tail
area being 0.025. The rejection region is bounded by critical values ±𝑧0.025 = ±1.95996.
Step 5: Now, we select the random sample from the population, collect the data, and do the
calculations. We find that 𝑦̅ 23.94 oz, s = 0.13 oz., z = -4.6154.
Step 6: We reject H0 at the 0.05 level of significance. We have sufficient evidence to conclude that
the mean amount of Post Grape-Nuts in a 24-oz. box is not 24 oz.
Next, we find a 95% confidence interval for the population mean, using the data from the sample.
The interval estimate is
𝑠
𝑦̅ ± 𝑧0.025
= (23.915 𝑜𝑧. , 23.965 𝑜𝑧. ).
√100
Since the interval does not include the target value of 24 oz., we conclude that the mean is not 24
oz.
Example 2: Hypothesis test about a population proportion.
We are given a coin and told that the coin may not be fair. We want to test this claim using
n = 1000 flips of the coin (I will leave it to the reader to collect the data).
Since we are testing hypotheses about a population proportion, we are collecting the data
using a binomial experiment. There are 1000 trials in the experiment. The trials are independent of
each other. The trials are identical to each other because each trial consists of flipping the coin and
recording whether the result is head (Success) or tail (failure). There are two possible outcomes for
each trial: either the result is head or the result is tail. Due to random sampling, the probability of
success is the same for each trial, namely the probability that a head results from a single flip of the
coin. We will conduct the test at the α = 0.01 level of significance. We flip the coin 1000 times and
find that we obtain 482 heads.
Step 1: H0: p = 0.50
Step 2: n = 1000,  = 0.01
versus
Step 3: The test statistic is Z 
Ha: p  0.59
pˆ  0.50
 0.50  0.50 
, which under H0 has an approximate standard
1000
normal distribution.
Step 4: The rejection region is bounded by the critical values -2.3263 and 2.3263.
Step 5: Now, we perform the experiment, collect the data, and do the calculations. We find that
482
pˆ 
 0.482 , z = -1.1384.
1000
Step 6: We fail to reject H0 at the 0.01 level of significance. We do not have sufficient evidence to
conclude that the coin is not fair.
Now we use the confidence interval approach, with a confidence level of 99%. The confidence
interval is
𝑝̂ (1 − 𝑝̂ )
𝑝̂ ± 𝑧0.005 √
= (0.445, 0.519).
1000
Since the interval does contain 0.50, we fail to reject the null hypothesis at the 0.01 level of
significance. We do not have sufficient evidence to conclude that the coin is not fair.