Standard +0.3 This is a straightforward chi-squared goodness of fit test with given proportions. Students must calculate expected frequencies (multiply proportions by 120), compute the test statistic using the standard formula, and compare to critical value at 10% significance with 4 degrees of freedom. It's slightly above average difficulty due to unequal proportions and requiring careful arithmetic, but follows a standard procedure with no conceptual challenges.
6 A large population of plants consists of five species \(A , B , C , D\) and \(E\) in the proportions \(p _ { A } , p _ { B } , p _ { C } , p _ { D }\) and \(p _ { E }\) respectively. A random sample of 120 plants consisted of \(23,14,24,27\) and 32 of \(A , B , C , D\) and \(E\) respectively. Carry out a test at the \(10 \%\) significance level of the null hypothesis that the proportions are \(p _ { \mathrm { A } } = p _ { \mathrm { B } } = 0.15 , p _ { \mathrm { C } } = p _ { \mathrm { D } } = 0.25\) and \(p _ { \mathrm { E } } = 0.2\).
\((H_0: p_A=p_B=0.15,\ p_c=p_D=0.25,\ p_E=0.20)\), \(H_1\): all alternatives
E values: \(18\ 18\ 30\ 30\ 24\)
B1
SC 5 difference of sample proportion from population proportion. \(z = 1.28, -1.02, 1.26, -0.63, 1.83\) B1
\(\chi^2 = 25/18+16/18+36/30+9/30+64/32 = 6.444\)
M1A1\(\sqrt{}\), A1
M1 for at least one correct, A1 for all correct ft E-values. Compare with \(-1.645\) AND \(1.645\) M1. E shoes a sig. diff., others don't. A1cwo
\((\alpha)\) CV \(= 7.779\)
B1
\(6.444 < CV\) so do not reject \(H_0\)
M1
\((\beta)\ P(\chi^2 > 6.444) = 0.168\)
B1
\(0.168 > 0.10\) so do not reject \(H_0\)
M1
There is insufficient evidence at the 10% significance level that the proportions are not as specified
A1
cwo, except rounding errors. NOT Insuff evidence that data does not fit the proportions. NOR Evidence that data fits the proportions.
[7]
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(H_0: p_A=p_B=0.15,\ p_c=p_D=0.25,\ p_E=0.20)$, $H_1$: all alternatives | | |
| E values: $18\ 18\ 30\ 30\ 24$ | B1 | SC 5 difference of sample proportion from population proportion. $z = 1.28, -1.02, 1.26, -0.63, 1.83$ B1 |
| $\chi^2 = 25/18+16/18+36/30+9/30+64/32 = 6.444$ | M1A1$\sqrt{}$, A1 | M1 for at least one correct, A1 for all correct ft E-values. Compare with $-1.645$ AND $1.645$ M1. E shoes a sig. diff., others don't. A1cwo |
| $(\alpha)$ CV $= 7.779$ | B1 | |
| $6.444 < CV$ so do not reject $H_0$ | M1 | |
| $(\beta)\ P(\chi^2 > 6.444) = 0.168$ | B1 | |
| $0.168 > 0.10$ so do not reject $H_0$ | M1 | |
| There is insufficient evidence at the 10% significance level that the proportions are not as specified | A1 | cwo, except rounding errors. NOT Insuff evidence that data does not fit the proportions. NOR Evidence that data fits the proportions. |
| | **[7]** | |
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6 A large population of plants consists of five species $A , B , C , D$ and $E$ in the proportions $p _ { A } , p _ { B } , p _ { C } , p _ { D }$ and $p _ { E }$ respectively. A random sample of 120 plants consisted of $23,14,24,27$ and 32 of $A , B , C , D$ and $E$ respectively. Carry out a test at the $10 \%$ significance level of the null hypothesis that the proportions are $p _ { \mathrm { A } } = p _ { \mathrm { B } } = 0.15 , p _ { \mathrm { C } } = p _ { \mathrm { D } } = 0.25$ and $p _ { \mathrm { E } } = 0.2$.
\hfill \mbox{\textit{OCR S3 2013 Q6 [7]}}