1. Quantitative Methods
Varsha Varde

2. Small-Sample Tests of Hypothesis
Contents:
1. Introduction
2. Student’s t distribution
3. Small-sample inferences about a population mean
4. Small-sample inferences about the difference between two means: Independent Samples
5. Small-sample inferences about the difference between two means: Paired Samples
6. Inferences about a population variance
7. Comparing two population variances
Varsha Varde

3. Introduction
When the sample size is small we only deal with normal populations.
For non-normal (e.g. binomial) populations different techniques are necessary
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4. Summary of Test Statistics to be Used in a Hypothesis Test about a Population Mean
Yes
n > 30 ? No
σ known ? No
Popul.
approx.
normal ?
Yes
Yes
Use s to
estimate σ
σ known ? No No
Yes
Use s to
Estimate σ
Increase n
to > 30

5. Student’s t Distribution
For small samples (n < 30) from normal populations, we have
z =x¯ - µ/σ/√n
If σ is unknown, we use s instead; but we no more have a Z distribution
Assumptions.
1. Sampled population is normal
2. Small random sample (n < 30)
3. σ is unknown
t =x¯ - µ/s/√n
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6. Properties of the t Distribution:
(i) It has n - 1 degrees of freedom (df)
(ii) Like the normal distribution it has a symmetric mound-shaped probability distribution
(iii) More variable (flat) than the normal distribution
(iv) The distribution depends on the degrees of freedom. Moreover, as n becomes larger, t converges to Z.
(v) Critical values (tail probabilities) are obtained from the t table
Examples.
(i) Find t0.05,5 = 2.015
(ii) Find t0.005,8 = 3.355
(iii) Find t0.025,26 = 2.056
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7. Small-Sample Inferences About a Population Mean
Parameter of interest: µ
Sample data: n, x¯, s
Other information: µ0= target value ;α
Point estimator: x¯
Estimator mean: µx¯ = µ
Estimated standard errorσx¯ = s/√n
Confidence Interval for µ:
x ± tα/2 ,n-1 (s/√n )
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8. Student’s t test Test: H0 : µ = µ0
Ha : 1) µ > µ0; 2) µ < µ0; 3) µ = µ0.
Critical value: either tα,n-1 or tα/2 ,n-1
T.S. : t = x¯-µ0/s/√n
RR:1) Reject H0 if t > tα,n-1
2) Reject H0 if t < -tα,n-1
3) Reject H0 if t > tα/2 ,n-1 or t < -tα/2,n-1
Decision: 1) if observed value is in RR: “Reject H0”
2) if observed value is not in RR: “Do not reject H0”
Conclusion: At 100α% significance level there is (in)sufficient statistical evidence to “favor Ha” .
Assumptions.1. Small sample (n < 30)
2. Sample is randomly selected
3. Normal population
4. Unknown variance
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9. Example
Example: It is claimed that weight loss in a new diet program is at least 20 pounds during the first month. Formulate &Test the appropriate hypothesis
Sample data: n = 25, x¯ =19.3, s2 = 25, µ0 = 20, α=0.05
Critical value: t0.05,24 =
H0 : µ ≥ 20 (µ is greater than or equal to 20 )
Ha : µ < 20,
T.S.:t =(x¯ - µ0 )/s/√n=(19.3 – 20)/5/√25 =-0.7
RR: Reject H0 if t <-1.711
Decision: Reject H0
Conclusion: At 5% significance level there is insufficient statistical evidence to conclude that weight loss in a new diet program exceeds 20 pounds per first month.
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10. Small-Sample Inferences About the Difference Between Two Means: Independent Samples
Parameter of interest: µ1 - µ2
Sample data:Sample 1: n1, x1, s1 ;Sample 2: n2, x2, s2
Other information: D0= target value ; α
Point estimator: X¯1 – X¯2
Estimator mean: µ X¯1 – X¯2 = µ1 - µ2
Assumptions.
1. Normal populations 2. Small samples ( n1 < 30; n2 < 30)
3. Samples are randomly selected4. Samples are independent
5. Variances are equal with common variance
σ2 = σ21 = σ22
Pooled estimator for σ.
s = √[(n1 - 1)s21 + (n2 - 1)s22]/(n1 + n2 – 2)
Estimator standard error:
σX¯1 – X¯2 = σ √ (1/n1+1/n2)
Confidence Interval:
(x¯1 - x¯2) ± (tα/2 ,n1+n2-2)(s √1/n1+1/n2)
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11. Two Means: Independent Samples
Test:H0 : µ1 - µ2 = D0
Ha : 1)µ1 - µ2 > D0 or
2) µ1 - µ2 < D0 or
3) µ1 - µ2 = D0
T.S. :t =(x¯1 - x¯2) - D0/s √1/n1+ 1/n2
RR: 1) Reject H0 if t > tα,n1+n2-2
2) Reject H0 if t < - tα,n1+n2-2
3) Reject H0 if t > tα/2,n1+n2-2 or t < - tα/2,n1+n2-2
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12. Example.(Comparison of two weight loss programs)
Refer to the weight loss example. Test the hypothesis that weight loss in a new diet program is different from that of an old program. We are told that that the observed value of Test Statistics is 2.2 and we know that
Sample 1 : n1 = 7 ; Sample 2 : n2 = 8 ; α= 0.05
Solution.
H0 : µ1 - µ2 = 0
Ha : µ1 - µ2 ≠0
T.S. :t =(x1 - x2) – 0/s √1/n1+ 1/n2= 2.2
Critical value: t.025,13 = 2.16
RR: Reject H0 if t > or t <
Decision: Reject H0
Conclusion: At 5% significance level there is sufficient statistical evidence to conclude that weight loss in the two diet programs are different
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13. Small-Sample Inferences About the Difference Between Two Means: Paired Samples
Parameter of interest: µ1 - µ2 = µd
Sample of paired differences data:
Sample : n = number of pairs, d¯ = sample mean, sd
Other information: D0= target value, α
Point estimator: d¯
Estimator mean: µd¯ = µd
Assumptions.
1. Normal populations
2. Small samples ( n1 < 30; n2 < 30)
3. Samples are randomly selected
4. Samples are paired (not independent)
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14. Sample standard deviation of the sample of n paired differences
sd = √ Σ (di - ¯d)2/ n-1
i=1
Estimator standard error: σd = sd/√n
Confidence Interval.
¯d ± tα/2,n-1sd/√n
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15. Test. H0 : µ1 - µ2 = D0 (equivalently, µd = D0)
Ha : 1) µ1 - µ2 = µd > D0;
2) µ1 - µ2 = µd < D0;
3) µ1 - µ2 = µd = D0,
T.S. :t =¯d - D0/sd/√n
RR:
1) Reject H0 if t > tα, n-1
2) Reject H0 if t < -tα,n-1
3) Reject H0 if t > tα/2,n-1 or t < -tα/2,n-1
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16. Example.
A manufacturer wishes to compare wearing qualities of two different types of tires, A and B. For the comparison a tire of type A and one of type B are randomly assigned and mounted on the rear wheels of each of five automobiles. The automobiles are then operated for a specified number of miles, and the amount of wear is recorded for each tire. These measurements are tabulated below.
Automobile Tire A Tire B
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17. Automobile Tire A Tire B d=A-B 1 10.6 10.2 0.4 2 9.8 9.4 0.4
x¯1 = x¯2 = 9.76
Using the previous section test we would have t = 0.57 resulting in an insignificant test which is inconsistent with the data.
Automobile Tire A Tire B d=A-B
x¯1 = x¯2 = d¯ =0 .48
Q1: Provide a summary of the data in the above table.
Sample summary: n = 5, d ¯= .48, sd = .0837
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18. Test. (parameter µd = µ1 - µ2) H0 : µd = 0 Ha : µd ≠ 0
Q2: Do the data provide sufficient evidence to indicate a difference in average wear for the two tire types.
Test. (parameter µd = µ1 - µ2)
H0 : µd = 0
Ha : µd ≠ 0
T.S. :t =d ¯- D0/sd/√n
=.48 – 0/.0837/√ = 12.8
RR: Reject H0 if t > or t < ( t.025,4 = 2.776)
Decision: Reject H0
Conclusion: At 5% significance level there is sufficient statistical evidence to to conclude that the average amount of wear for type A tire is different from that for type B tire.
Exercise. Construct a 99% confidence interval for the difference in average wear for the two tire types.
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19. Inferences About a Population Variance
Chi-square distribution. When a random sample of size n is drawn from a normal population with mean µ and standard deviation σ, the sampling distribution of S2 depends on n. The standardized distribution of S2 is called the chi-square distribution and is given by
χ2 =(n - 1)s2/σ2
Degrees of freedom (df): ν = n - 1
Graph: Non-symmetrical and depends on df
Critical values: using χ2 tables
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20. Inferences About a Population Variance
Test.H0 : σ2 = σ20
Ha : σ2 ≠ σ20 (two-tailed test).
Ha : σ2 <σ20 (One –tailed test )
T.S. :X2 =(n - 1)s2/ σ20
RR: (two-tailed test).
Reject H0 if X2 > X2α/2 or X2 < X2 1-α/2 where
X2 is based on (n - 1) degrees of freedom.
RR: (One-tailed test).
Reject H0 if X2 < X2 1-α where
X2 is based on (n - 1) degrees of freedom
Assumptions.
1. Normal population
2. Random sample
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21. Example
Future Technologies ltd. manufactures high resolution telescopes .The management wants its products to have a variation of less than 2 sd in resolution while focusing on objects which are beyond 500 light years. When they tested their newly manufactured telescope for 30 times to focus on an object 500 light years away they found sample sd to be State the Hypothesis and test it at 1% level of significance. Can the management accept to sell this product?
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22. Solution Given ,n=30 and S2=(1.46)2, σ20 =4 We set up hypothesis
Ha: σ2< 4
Significance Level α =1%=.01
This is a one tailed test
T.S = X2 =(n - 1)s2/ σ20= (30-1)(1.46)2/4=15.45
Referring to X2 tables we find at 29 d.f the value of X2 that leaves an area of =.99 in the upper tail and .01 in the lower tail is Since calculated value is greater than the cut off level we accept null hypothesis and conclude that sd=2. There fore management will not allow sale of its telescope.
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23. Comparing Two Population Variances
F-distribution. When independent samples are drawn from two normal populations with equal variances then S21/S22 possesses a sampling distribution that is known as an
F distribution. That is
F = S21/S22
Degrees of freedom (df): ν1 = n1 - 1; ν2 = n2 - 1
Graph: Non-symmetrical and depends on df
Critical values: using F tables
Test.
H0 : σ21 = σ22
Ha : σ21≠ σ22(two-tailed test).
T.S. :F = S21/S22
Where S21is the larger sample variance.
Note: F = larger sample variance/smaller sample variance
RR: Reject H0 if F > Fα/2 where Fα/2 is based on (n1 - 1) and (n2 - 1) degrees of freedom.
Assumptions.
1. Normal populations 2. Independent random samples
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24. Example Investment 1: n1 = 10, x¯1 = 17.8%; s21 = 3.21
Investment risk is generally measured by the volatility of possible outcomes of the investment. The most common method for measuring investment volatility is by computing the variance ( or standard deviation) of possible outcomes. Returns over the past 10 years for first alternative and 8 years for the second alternative produced the following data:
Data Summary:
Investment 1: n1 = 10, x¯1 = 17.8%; s21 = 3.21
Investment 2: n2 = 8, x¯2 = 17.8%; s22 = 7.14
Both populations are assumed to be normally distributed.
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25. Solution
Q1: Do the data present suffcient evidence to indicate that the risks for investments1 and 2 are unequal ?
Solution.
Test:H0 : σ21 = σ22
Ha : σ21 ≠ σ22 (two-tailed test).
T.S. :F = S21/S22 =7.14/3.21= 2.22
.RR: Reject H0 if F > Fα/2 where
Fα/2,n2-1,n1-1 = F.025,7,9 = 4.20
Decision: Do not reject H0
Conclusion: At 5% significance level there is insuffcient statistical evidence to indicate that the risks for investments 1 and 2 are unequal.
Exercise. Do the upper tail test. That is
Ha : σ21 > σ22.
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