Standard +0.3 This is a straightforward one-sample z-test with known variance (assumed from population), requiring calculation of sample mean, test statistic, and comparison with critical value. While it's a 10-mark question requiring detailed working, it follows a standard hypothesis testing template with no conceptual surprises—slightly easier than average since it's purely procedural application of a well-rehearsed method.
8 In this question you must show detailed reasoning.
On the manufacturer's website, it is claimed that the average daily electricity consumption of a particular model of fridge is 1.25 kWh (kilowatt hours). A researcher at a consumer organisation decides to check this figure.
A random sample of 40 fridges is selected. Summary statistics for the electricity consumption \(x \mathrm { kWh }\) of these fridges, measured over a period of 24 hours, are as follows.
\(\Sigma x = 51.92 \quad \Sigma x ^ { 2 } = 70.57\)
Carry out a test at the \(5 \%\) significance level to investigate the validity of the claim on the website. [0pt]
[10]
1.064 < 1.96 so not significant (do not reject H )
0
Insufficient evidence to suggest that the website claim
Answer
Marks
is not true
B1
M1
A1
B1
B1
B1
M1
B1
M1
A1
Answer
Marks
[10]
1.1
1.1
1.1
3.4
1.1a
1.2
3.3
1.1
2.2b
Answer
Marks
3.5a
DR
1
= ×3.1778
39
or est. pop sd = 0.2854
soi
Both correct; if stated in words only,
must include ‘population’
For definition of µ in context
Allow wrong mean or sd here
FT wrong (sensible) c.v. and test
statistic if calculation is of right form
Answer
Marks
for M1 but not for A1
Alternative: calculate
P(X > 1.298) using
N(1.25, 0.0815/40)
= 0.856
Use of 0.975 (not 0.95)
Question 8:
8 | 51.92
Est. population mean = =1.298
40
1 ( 51.922)
Est. population variance = 70.57−
39 40
= 0.0815
Test is based on a Normal distribution
H : µ = 1.25, H : µ ≠ 1.25,
0 1
where µ is the population mean (electricity
consumption)
1.298−1.25
Test statistic is
0.0815/40
= 1.064
Critical value (2-tailed) at 5% level is 1.96
1.064 < 1.96 so not significant (do not reject H )
0
Insufficient evidence to suggest that the website claim
is not true | B1
M1
A1
B1
B1
B1
M1
B1
M1
A1
[10] | 1.1
1.1
1.1
3.4
1.1a
1.2
3.3
1.1
2.2b
3.5a | DR
1
= ×3.1778
39
or est. pop sd = 0.2854
soi
Both correct; if stated in words only,
must include ‘population’
For definition of µ in context
Allow wrong mean or sd here
FT wrong (sensible) c.v. and test
statistic if calculation is of right form
for M1 but not for A1 | Alternative: calculate
P(X > 1.298) using
N(1.25, 0.0815/40)
= 0.856
Use of 0.975 (not 0.95)
8 In this question you must show detailed reasoning.
On the manufacturer's website, it is claimed that the average daily electricity consumption of a particular model of fridge is 1.25 kWh (kilowatt hours). A researcher at a consumer organisation decides to check this figure.
A random sample of 40 fridges is selected. Summary statistics for the electricity consumption $x \mathrm { kWh }$ of these fridges, measured over a period of 24 hours, are as follows.\\
$\Sigma x = 51.92 \quad \Sigma x ^ { 2 } = 70.57$
Carry out a test at the $5 \%$ significance level to investigate the validity of the claim on the website.\\[0pt]
[10]
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q8 [10]}}