AQA S3 2009 June — Question 4 8 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2009
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample z-test large samples
DifficultyStandard +0.8 This is a two-sample t-test requiring students to formulate hypotheses for a difference claim (μ₁ - μ₂ > 15, not just μ₁ > μ₂), calculate pooled variance, compute the test statistic, and interpret at 1% significance. The non-zero difference in the null hypothesis and large sample sizes add complexity beyond standard two-sample tests, but it remains a structured application of learned procedures.
Spec5.05c Hypothesis test: normal distribution for population mean

4 Holly, a horticultural researcher, believes that the mean height of stems on Tahiti daffodils exceeds that on Jetfire daffodils by more than 15 cm . She measures the heights, \(x\) centimetres, of stems on a random sample of 65 Tahiti daffodils and finds that their mean, \(\bar { x }\), is 40.7 and that their standard deviation, \(s _ { x }\), is 3.4 . She also measures the heights, \(y\) centimetres, of stems on a random sample of 75 Jetfire daffodils and finds that their mean, \(\bar { y }\), is 24.4 and that their standard deviation, \(s _ { y }\), is 2.8 . Investigate, at the \(1 \%\) level of significance, Holly's belief.

Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(H_0: \mu_X - \mu_Y = 15\)B1 Or equivalent; accept \(H_0: \mu_X - \mu_Y = 0\)
\(H_1: \mu_X - \mu_Y > 15\)B1 Or equivalent
CV \(z = 2.32\) to \(2.33\)B1 AWFW (2.3263); if \(H_1\) involves \(\neq\) accept 2.57 to 2.58 (2.5758)
CV \(t = 2.35\) to \(2.36\)(B1) AWFW; if \(H_1\) involves \(\neq\) accept 2.60 to 2.62
\(z = \dfrac{(\bar{x}-\bar{y})-15}{\sqrt{\frac{s_X^2}{n_X}+\frac{s_Y^2}{n_Y}}}\) or \(z/t = \dfrac{(\bar{x}-\bar{y})-15}{\sqrt{s_p^2\left(\frac{1}{n_X}+\frac{1}{n_Y}\right)}}\)M1 Used; allow 'no \(-15\)'
\(s_p^2 = \dfrac{(64\times3.4^2)+(74\times2.8^2)}{65+75-2} = \dfrac{1320}{138} = 9.56522\) \(s_p = 3.09277\)
\(z = \dfrac{(40.7-24.4)-15}{\sqrt{\frac{3.4^2}{65}+\frac{2.8^2}{75}}} = \dfrac{1.3}{\sqrt{0.28238}}\)A1 Numerator; allow 'no \(-15\)'
A1Denominator
\(= 2.44\) to \(2.45\)A1 AWFW (2.4464); 'no \(-15\)' gives \(z=30.674\)
OR (pooled): \(z/t = \dfrac{(40.7-24.4)-15}{\sqrt{\frac{1320}{138}\left(\frac{1}{65}+\frac{1}{75}\right)}} = \dfrac{1.3}{\sqrt{0.27469}}\)(A1),(A1) Numerator; Denominator
\(= 2.48\)(A1) AWRT (2.4804); 'no \(-15\)' gives \(z=31.100\)
Thus evidence, at 1% level, to support Holly's beliefA1F 8
# Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_X - \mu_Y = 15$ | B1 | Or equivalent; accept $H_0: \mu_X - \mu_Y = 0$ |
| $H_1: \mu_X - \mu_Y > 15$ | B1 | Or equivalent |
| CV $z = 2.32$ to $2.33$ | B1 | AWFW (2.3263); if $H_1$ involves $\neq$ accept 2.57 to 2.58 (2.5758) |
| CV $t = 2.35$ to $2.36$ | (B1) | AWFW; if $H_1$ involves $\neq$ accept 2.60 to 2.62 |
| $z = \dfrac{(\bar{x}-\bar{y})-15}{\sqrt{\frac{s_X^2}{n_X}+\frac{s_Y^2}{n_Y}}}$ or $z/t = \dfrac{(\bar{x}-\bar{y})-15}{\sqrt{s_p^2\left(\frac{1}{n_X}+\frac{1}{n_Y}\right)}}$ | M1 | Used; allow 'no $-15$' |
| $s_p^2 = \dfrac{(64\times3.4^2)+(74\times2.8^2)}{65+75-2} = \dfrac{1320}{138} = 9.56522$ | | $s_p = 3.09277$ |
| $z = \dfrac{(40.7-24.4)-15}{\sqrt{\frac{3.4^2}{65}+\frac{2.8^2}{75}}} = \dfrac{1.3}{\sqrt{0.28238}}$ | A1 | Numerator; allow 'no $-15$' |
| | A1 | Denominator |
| $= 2.44$ to $2.45$ | A1 | AWFW (2.4464); 'no $-15$' gives $z=30.674$ |
| **OR** (pooled): $z/t = \dfrac{(40.7-24.4)-15}{\sqrt{\frac{1320}{138}\left(\frac{1}{65}+\frac{1}{75}\right)}} = \dfrac{1.3}{\sqrt{0.27469}}$ | (A1),(A1) | Numerator; Denominator |
| $= 2.48$ | (A1) | AWRT (2.4804); 'no $-15$' gives $z=31.100$ |
| Thus evidence, at 1% level, to support Holly's belief | A1F | 8 | F on $z$ and CV |

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4 Holly, a horticultural researcher, believes that the mean height of stems on Tahiti daffodils exceeds that on Jetfire daffodils by more than 15 cm .

She measures the heights, $x$ centimetres, of stems on a random sample of 65 Tahiti daffodils and finds that their mean, $\bar { x }$, is 40.7 and that their standard deviation, $s _ { x }$, is 3.4 .

She also measures the heights, $y$ centimetres, of stems on a random sample of 75 Jetfire daffodils and finds that their mean, $\bar { y }$, is 24.4 and that their standard deviation, $s _ { y }$, is 2.8 .

Investigate, at the $1 \%$ level of significance, Holly's belief.

\hfill \mbox{\textit{AQA S3 2009 Q4 [8]}}