Moderate -0.3 This is a straightforward application of a one-sample t-test with all necessary summary statistics provided. Students must calculate the sample standard deviation, compute the t-statistic, compare to critical values, and state a conclusion. While it requires careful execution of the standard procedure, it involves no conceptual challenges or novel problem-solving—just routine application of a textbook method with clearly defined hypotheses.
1 The heights of the members of a large sports club are normally distributed. A random sample of 11 members of the club is chosen and their heights, \(x \mathrm {~cm}\), are measured. The results are summarised as follows, where \(\bar { x }\) denotes the sample mean of \(x\).
$$\bar { x } = 176.2 \quad \sum ( x - \bar { x } ) ^ { 2 } = 313.1$$
Test, at the \(5 \%\) significance level, the null hypothesis that the population mean height for members of this club is equal to 172.5 cm against the alternative hypothesis that the mean differs from 172.5 cm . [5]
Accept \(H_0\): insufficient evidence to reject mean height is \(172.5\), OR sufficient evidence to accept mean height is \(172.5\)
A1 FT
Correct conclusion in context, level of uncertainty in language used. No contradictions. CWO. Do not accept: 'mean height is 172.5'. Do not accept use of symbols that are undefined e.g. insufficient evidence to reject \(\mu = 172.5\)
5
## Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2 = \frac{313.1}{10} = 31.3$ | **B1** | Can be implied |
| $t = \frac{176.2 - 172.5}{s / \sqrt{11}}$ | **M1** | |
| $= 2.19$ | **A1** | |
| $'2.19' < 2.228$ | **M1** | Comparison of their $t$ value with $2.228$ |
| Accept $H_0$: insufficient evidence to reject mean height is $172.5$, OR sufficient evidence to accept mean height is $172.5$ | **A1 FT** | Correct conclusion in context, level of uncertainty in language used. No contradictions. CWO. Do not accept: 'mean height is 172.5'. Do not accept use of symbols that are undefined e.g. insufficient evidence to reject $\mu = 172.5$ |
| | **5** | |
1 The heights of the members of a large sports club are normally distributed. A random sample of 11 members of the club is chosen and their heights, $x \mathrm {~cm}$, are measured. The results are summarised as follows, where $\bar { x }$ denotes the sample mean of $x$.
$$\bar { x } = 176.2 \quad \sum ( x - \bar { x } ) ^ { 2 } = 313.1$$
Test, at the $5 \%$ significance level, the null hypothesis that the population mean height for members of this club is equal to 172.5 cm against the alternative hypothesis that the mean differs from 172.5 cm . [5]\\
\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q1 [5]}}