BMMT120 — Tutorial Sheet 7 (Solutions)
Prepared for: University of Lusaka
Mr. M. Chipoka & Ms. A. Kawama (questions)
October 7, 2025
Notation: For sample of size n, sample mean x̄ = n1
Tests are at the 5% level unless otherwise stated.
P
1
xi , sample variance s2 = n−1
P
x2i = 1693.6331.
P
Question 1
A random sample x1 , . . . , x10 gives
P
xi = 120.19 and
(a)(i) Sample mean and variance
n = 10,
x̄ =
120.19
= 12.019.
10
Sample variance
P
1 X 2 ( xi )2 1 (120.19)2 s =
xi −
=
1693.6331 −
≈ 27.6744,
n−1
n
9
10
so s ≈ 5.2606.
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(a)(ii) Test: H0 : µ = 15 (population variance unknown)
Use the one-sample t statistic:
t=
x̄ − 15
12.019 − 15
√ =
√ ≈ −1.7919.
s/ n
5.2606/ 10
Degrees of freedom: ν = 9. Two-sided p-value:
p ≈ 0.1067.
Conclusion: At the 5% level we fail to reject H0 (no significant evidence that the population
mean differs from 15 when variance is unknown).
(a)(iii) Test: H0 : µ = 15 when population variance is known (σ 2 = 20)
If σ 2 = 20 then σ =
√
20 and we use the Z-statistic:
x̄ − 15
12.019 − 15
√ = √
√
Z=
≈ −2.1079,
σ/ n
20/ 10
Two-sided p-value ≈ 0.0350. Conclusion: At 5% level we reject H0 (so with known variance 20
the sample gives evidence that the mean differs from 15).
1
xi )2
n
P
x2i − (
.
Question 1(b)
Two independent samples from normal populations, assumed equal population variance:
s21 = 12.4,
s22 = 25.8,
n1 = 10;
n2 = 5,
sample means differ by ∆x̄ = 4.5.
(i) Test for equality of population means
When variances are assumed equal we use the pooled t:
s2p =
(n1 − 1)s21 + (n2 − 1)s22
9 · 12.4 + 4 · 25.8
=
n1 + n 2 − 2
13
Compute:
s2p ≈
214.8
111.6 + 103.2
=
≈ 16.5231,
13
13
sp ≈ 4.0646.
Test statistic
t=
x̄ − x̄2
4.5
4.5
4.5
q1
q
√
√
=
≈
=
≈ 1.958.
1
1
1
1
4.0646
0.1
+
0.2
4.0646 0.3
sp n1 + n2
4.0646 10 + 5
Degrees of freedom ν = n1 + n2 − 2 = 13. Two-sided p-value ≈ 0.074 (approx). Conclusion: At
5% level we fail to reject equality of means (no strong evidence of a difference).
(ii) Sensitivity to equal-variance assumption
If the equal-variance assumption is violated the pooled t is not valid (risk of incorrect Type I error).
In that case use the Welch t-test (unequal-variance t) which adjusts the standard error and uses
Welch–Satterthwaite df. The Welch test is robust to unequal variances and is the recommended
alternative.
Question 2
Treatment A:
{24.5, 21.7, 35.2, 15.9, 23.7, 34.2, 29.3, 21.1, 23.5, 28.3}
Treatment B:
{22.4, 21.2, 36.3, 15.7, 21.5, 7.3, 12.8, 21.2, 23.9, 18.4}
Let nA = nB = 10.
(a) Test whether variances are equal
Sample variances (ddof=1):
s2A ≈ 36.1871,
s2B ≈ 58.4357.
s2A
≈ 0.6193 (we may place larger variance in numerator for a one-sided
s2B
test). Two-sided p-value ≈ 0.4864. Conclusion: At the 5% level we fail to reject equality of
variances.
Use the F statistic F =
2
(b) t-test for difference in means
Sample means:
x̄A ≈ 25.7400,
x̄B ≈ 20.0700,
Welch two-sample t statistic (unequal-variance) was computed as
t ≈ 1.8433,
two-sided p ≈ 0.0827.
Conclusion: At 5% level we fail to reject equality of means (evidence of a difference is weak,
p ≈ 0.083). Because the F-test did not show strong evidence of unequal variances, a pooled-t would
give a similar conclusion.
(c) Policy implications
• There is no strong statistical evidence that mean reductions differ between treatments A and
B at 5% (though mean for A is larger numerically).
• Decision-makers might consider sample size: larger trials could detect smaller effect sizes.
• If Treatment B is cheaper or safer but not significantly different in mean reduction, policy
could prefer B; otherwise keep A if clinical or economic reasons support it.
Question 3
For a sample of hospitals:
n
P
P xi2
xi
Private
35
13,125
5,250,000
Public
61
14,640
4,050,000
(a) Means and variances
Private:
13125
= 375.0.
35
1
(13,125)2 5,250,000 −
≈ 9650.7353.
s21 =
34
35
x̄1 =
Public:
14,640
= 240.0.
61
1
(14,640)2 s22 =
4,050,000 −
= 8940.0.
60
61
x̄2 =
(b) Test whether private hospitals have greater variance (one-sided)
Test H0 : σ12 ≤ σ22 vs Ha : σ12 > σ22 using
F =
s21
9650.7353
≈
≈ 1.0797.
2
8940.0
s2
Given the critical value F0.05;34,60 = 1.56, since F < 1.56 we fail to reject H0 at 5% (no evidence
that private variance is greater).
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(c) Test whether means differ
Observed mean difference x̄1 − x̄2 = 375 − 240 = 135.
Pooled standard deviation (since variances are similar and F was not significant):
s
(n1 − 1)s21 + (n2 − 1)s22
sp =
≈ 95.9014.
n1 + n 2 − 2
Test statistic
t=
135
q
≈ 6.6385,
1
1
sp 35
+ 61
df = 35 + 61 − 2 = 94, two-sided p ≈ 2.02 × 10−9 . Conclusion: Very strong evidence that mean
expenditure in private hospitals exceeds that in public hospitals.
(d) Implications and policy
• Private hospitals have a significantly higher mean expenditure per patient visit (here x̄1 =
K375 vs x̄2 = K240).
• Policy consideration: investigate causes (service level, pricing, case-mix). If equity of access is
a goal, consider subsidies or price regulation for vulnerable groups, or encourage competition
/ regulation to reduce unwarranted cost differences.
Question 4
Male nurses: nm = 90, x̄m = 25.7, sm = 8.2.
Female nurses: nf = 80, x̄f = 22.5, sf = 3.5.
(a) Test whether male mean exceeds female mean by more than one year
Null/alternative:
H0 : µm − µf = 1,
Ha : µm − µf > 1.
Observed difference: d = x̄m − x̄f = 3.20.
Standard error:
s
r
s2f
s2m
8.22 3.52
SE =
=
+
+
≈ 0.9488.
nm nf
90
80
Test statistic
d−1
3.20 − 1
=
≈ 2.3187.
SE
0.9488
Degrees of freedom by Welch approximation ν ≈ 123.4. One-sided p ≈ 0.0110. Conclusion: At
5% level we reject H0 and conclude that male nurses have on average more than one year greater
experience than female nurses.
t=
(b) Policy implications
• If male nurses have significantly more experience, consider recruitment or retention policies
focused on balance and training opportunities for female nurses.
• Investigate causes (historical hiring, career interruptions, promotion pathways) to address
inequality if desired.
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Question 5
Attendance for fee K170: {115, 110, 125, 120, 120, 105, 115, 125, 115, 120}.
Attendance for fee K200: {105, 100, 95, 110, 105, 100, 95, 100, 105, 93}.
Both samples size n = 10.
(a) Means and standard deviations
x̄170 = 117.0,
s170 ≈ 6.3246.
x̄200 = 100.8,
s200 ≈ 5.4528.
(b) Test whether variances are equal (F-test)
F =
s2170
≈ 1.3453,
s2200
two-sided p ≈ 0.666.
Conclusion: Fail to reject equality of variances at 5%.
(c) Test whether mean daily attendance differs
Two-sample t test (Welch):
t ≈ 6.1347,
two-sided p ≈ 9.41 × 10−6 .
Conclusion: Strong evidence that mean attendance differs between fees (attendance is higher at
K170).
(d) 95% CI for difference in mean daily revenue (K170 revenue minus K200
revenue)
Compute daily revenue per day arrays:
Revenue170,i = 170 × (attendance at K170 on day i),
Revenue200,i = 200 × (attendance at K200 on day i).
Sample mean daily revenues:
R170 = K19,890.0,
R200 = K20,160.0.
Difference R170 − R200 = −K270.0 (i.e. K200 yields a slightly higher mean revenue).
95% CI (Welch-style) for R170 − R200 :
CI ≈ (−K1287.46, K747.46).
This interval contains 0, so the revenue difference is not statistically significant at the 5% level.
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(e) Recommendation
• Although K170 gives significantly higher attendance, K200 yields slightly higher mean revenue
(not statistically significant).
• Since revenue difference is inconclusive, choose on operational grounds: if higher attendance
burdens staff/resources or impacts quality, prefer K200; if aiming to maximize access and
total patients seen, prefer K170.
Notes on computations: Numerical values (means, variances, test statistics, p-values and confidence intervals) were computed to at least 3 significant digits where relevant. Tests used t- or
Z-distributions as appropriate; Welch’s adjustment applied in unequal-variance comparisons.
Source: Tutorial sheet uploaded (BMMT120 Tutorial Sheet 7, Oct 7, 2025).
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