Challenging +1.2 This is a standard two-sample t-test with unknown variances requiring calculation of the pooled variance, test statistic, and working backwards from a non-rejection decision to find the significance level range. While it involves multiple steps and reverse reasoning (finding α rather than making a decision given α), the procedure is mechanical and follows a well-practiced template from Further Maths statistics with no conceptual surprises.
A random sample of 50 observations of a random variable \(X\) and a random sample of 60 observations of a random variable \(Y\) are taken. The results for the sample means, \(\bar{x}\) and \(\bar{y}\), and the unbiased estimates for the population variances, \(s_x^2\) and \(s_y^2\), respectively, are as follows.
$$\bar{x} = 25.4 \quad \bar{y} = 23.6 \quad s_x^2 = 23.2 \quad s_y^2 = 27.8$$
A test, at the \(\alpha\%\) significance level, of the null hypothesis that the population means of \(X\) and \(Y\) are equal against the alternative hypothesis that they are not equal is carried out. Given that the null hypothesis is not rejected, find the set of possible values of \(\alpha\). [5]
Calculate value of z (to 2 d.p., either sign): z = = 1⋅869 M1 A1
s
Find Φ(z) and set of possible values of α (to 1 d.p.):
Φ(z) = 0⋅9692 [or 96⋅92%]
Answer
Marks
Guidance
(M1 A0 for α < 3⋅1 or α > 93⋅8) α < (or <) 6⋅2 (allow 6⋅1) M1 A1
5
[5]
Page 8
Mark Scheme
Syllabus
Cambridge International A Level – October/November 2014
9231
22
S.R. Assuming equal population variances: Explicit assumption (B1)
(49s 2 +59s 2)
Find pooled estimate of common variance s 2 X Y
108
2777
= or 25⋅71 or 5⋅071 2
108
1.8
and calculate value of z (to 2 d.p.): z = = 1⋅854 (M1 A1)
s 50−1+60−1
Find Φ(z) and values of α (to 1 d.p.): Φ(z) = 0⋅9681 [or 96⋅81%]
(M1 A0 for α < 3⋅2 or α > 93⋅6) α < (or <) 6⋅4 (M1 A1)
Question 6:
6 | Estimate population variance for combined sample:
s 2 s 2
s 2 = X + Y
50 60
1391
= or 0⋅9273 or 0⋅9630 2 M1
1500
1.8
Calculate value of z (to 2 d.p., either sign): z = = 1⋅869 M1 A1
s
Find Φ(z) and set of possible values of α (to 1 d.p.):
Φ(z) = 0⋅9692 [or 96⋅92%]
(M1 A0 for α < 3⋅1 or α > 93⋅8) α < (or <) 6⋅2 (allow 6⋅1) M1 A1 | 5 | [5]
Page 8 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
S.R. Assuming equal population variances: Explicit assumption (B1)
(49s 2 +59s 2)
Find pooled estimate of common variance s 2 X Y
108
2777
= or 25⋅71 or 5⋅071 2
108
1.8
and calculate value of z (to 2 d.p.): z = = 1⋅854 (M1 A1)
s 50−1+60−1
Find Φ(z) and values of α (to 1 d.p.): Φ(z) = 0⋅9681 [or 96⋅81%]
(M1 A0 for α < 3⋅2 or α > 93⋅6) α < (or <) 6⋅4 (M1 A1)
A random sample of 50 observations of a random variable $X$ and a random sample of 60 observations of a random variable $Y$ are taken. The results for the sample means, $\bar{x}$ and $\bar{y}$, and the unbiased estimates for the population variances, $s_x^2$ and $s_y^2$, respectively, are as follows.
$$\bar{x} = 25.4 \quad \bar{y} = 23.6 \quad s_x^2 = 23.2 \quad s_y^2 = 27.8$$
A test, at the $\alpha\%$ significance level, of the null hypothesis that the population means of $X$ and $Y$ are equal against the alternative hypothesis that they are not equal is carried out. Given that the null hypothesis is not rejected, find the set of possible values of $\alpha$. [5]
\hfill \mbox{\textit{CAIE FP2 2014 Q6 [5]}}