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ONE SAMPLE t-TEST FOR THE MEAN OF THE NORMAL
DISTRIBUTION
Let X 1 , , X n sample from N(μ, σ), μ and σ unknown , estimate σ using s.
Let significance level =α.
STEP 1. Ho: μ = μo
Ha: μ ≠ μo or
Ha: μ > μo or Ha: μ < μo
STEP 2. Compute the test statistic: t  x  o
s/ n
STEP 3. Compute the critical number/value
depends on Ha.
Two sided alternative
critical value = tα/2(n-1).
One sided alternative
critical value = tα(n-1).
STEP 4. DECISION-critical/rejection regions, use t distribution with df=n-1.
Ha: μ ≠ μo Reject Ho if |t|> tα/2(n-1);
Ha: μ > μo Reject Ho if t > tα(n-1);
Ha: μ < μo
Reject Ho if t < - tα(n-1).
STEP 5. Answer the question in the problem.
P-value approach
STEP 3. Compute the p-value.
Two sided test p-value:
Ha: μ ≠ μo, P-value: 2P( t(n-1)> |t|)
One sided tests p-values:
Ha: μ > μo, P-value: P( t(n-1) > t)
Ha: μ < μo, P-value: P( t(n-1) < t)
value of a
t-random variable
STEP 4. DECISION
Reject Ho if p-value < significance level α.
STEP 5. Answer the question in the problem.
value of the
test statistic
EXAMPLES
Example1. A sample of 36 women resulted in mean height of 64” and sample
variance = 25. Are women, on average, shorter than 66”? Use 5% significance
level. Assume heights follow normal distribution.
Solution. x  64, n=36, s2=25, μo=66, α = 5%.
STEP1. Ho: μ = 66
STEP 2. Test statistic t 
Ha: μ < 66
x  66 64  66

 2.4.
5/6
5 / 36
STEP3. Critical value? Df=n-1=35 and t0.05(35) ≈ 1.69 (Table C).
STEP 4. DECISION: t = -2.4 < -t0.05(35) ≈ -1.69
STEP5. On average, women are shorter than 66”.
reject Ho.
EXAMPLE 1 contd.
P-value approach
STEP 3.
P-value =
P(t (35)  2.4)  0.01
STEP 4. DECISION: P-value=0.01 < sign. level = 0.05
STEP5. On average, women are shorter than 66”.
reject Ho.
EXAMPLES, contd.
Example 2. Suppose a sample of size 16 results in mean of 10 and
standard deviation of 3.2. Test Ho: μ = 8 vs.
Ha: μ > 8 using
α=0.05.
Solution. x  10, n=16, s=3.2, μo=8, α = 5%.
STEP1. Ho: μ = 8
STEP 2. Test statistic
Ha: μ > 8
10  8
t
 2.5.
3.2 / 16
STEP3. Critical value? Df=n-1=15 and t0.05(15) ≈ 1.753 (Table C).
STEP 4. DECISION: t = 2.5 > t0.05(15)=1.753, so we reject Ho.
STEP5. Reject Ho.
EXAMPLES, contd.
Example 3. Suppose a sample of size 16 results in mean of 10 and
standard deviation of 3.2. Test Ho: μ = 8 vs. Ha: μ ≠ 8 using α=0.05.
Solution.
x  10,
n=16, s=3.2, μo=8, α = 5%.
STEP1. Ho: μ = 8
Ha: μ ≠ 8
STEP 2. Test statistic = 2.5.
STEP3. Critical value? Df=n-1=15, α/2=0.025, and t0.025(15) =2.131.
STEP 4. DECISION: t = 2.5 > t0.025(15)=2.131, so we reject Ho.
STEP5. Reject Ho.
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