It is believed that the probability pp of a randomly chosen pregnant woman giving birth to a Down's Syndrome child is related to the women's age xx, in years. The table gives the observed values of pp for 5 different values of xx.
xx2530354045
pp0.000670.001250.003330.010000.03330

  1. (i)

    Sketch the scatter diagram representing the data and comment if p=mx+cp = mx + c is a good model for this set of data.

  2. (ii)

    By finding the product-moment correlation coefficient for each of the following models, determine which of the following would be an appropriate model to represent the above data:

    A: p=a+bxp = a + \frac{b}{x},

    B: p=a+be−xp = a + b e^{-x},

    C: ln⁡p=a+bx\ln p = a + bx,

    where aa and bb are constants.

  3. (iii)

    For the appropriate model, calculate the values of aa and bb.

  4. (iv)

    Obtain an estimate of the probability of a 19-year-old woman giving birth to a Down's Syndrome child. Comment on the reliability of your answer.

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Worked solution

9(i)

1. Sketch the scatter diagram for the given data

[Sketch of a scatter diagram plotting the given points (x,p)(x, p) with xx on the horizontal axis and pp on the vertical axis. The points (25, 0.00067), (30, 0.00125), (35, 0.00333), (40, 0.01000) and (45, 0.03330) are plotted, clearly showing an upward-bending curve.]

2. Comment on the appropriateness of the linear model

Since the points on the scatter diagram exhibit a non-linear trend and follow a curve rather than a straight line, the linear model p=mx+cp = mx + c is not a good model for this set of data.

Answer

[Sketch of a scatter diagram with points curving upwards]

No, p=mx+cp = mx + c is not a good model as the points lie on a curve, not a straight line.

9(ii)

1. Calculate the product-moment correlation coefficient for Model A

For Model A (p=a+bxp = a + \frac{b}{x}), the linear relationship is between 1x\frac{1}{x} and pp.
By entering the data pairs (1x,p)\left(\frac{1}{x}, p\right) into a graphing calculator, we obtain the product-moment correlation coefficient:
r≈−0.7628r \approx -0.7628

2. Calculate the product-moment correlation coefficient for Model B

For Model B (p=a+be−xp = a + b e^{-x}), the linear relationship is between e−xe^{-x} and pp.
Entering the data pairs (e−x,p)(e^{-x}, p) into the graphing calculator gives:
r≈−0.3719r \approx -0.3719

3. Calculate the product-moment correlation coefficient for Model C

For Model C (ln⁡p=a+bx\ln p = a + bx), the relationship is directly linear between xx and ln⁡p\ln p.
Entering the data pairs (x,ln⁡p)(x, \ln p) into the graphing calculator yields:
r≈0.9937r \approx 0.9937

4. Determine the most appropriate model based on the correlation coefficients

To determine the best fit, we compare the absolute values of the correlation coefficients (∣r∣|r|).
Model C has a correlation coefficient magnitude closest to 11 (∣0.9937∣≈1|0.9937| \approx 1), which indicates the strongest linear correlation among the three models.

Thus, Model C is the most appropriate model.

Answer
  • Model A: r≈−0.7628r \approx -0.7628
  • Model B: r≈−0.3719r \approx -0.3719
  • Model C: r≈0.9937r \approx 0.9937

Model C is the most appropriate model.

9(iii)

1. Identify the appropriate model to use based on the expected results from part (ii).

From part (ii), Model C (ln⁡p=a+bx\ln p = a + bx) yields a product-moment correlation coefficient closest to 11 (r≈0.994r \approx 0.994), making it the most appropriate model for the data.

2. Use a graphing calculator to compute the linear regression coefficients for the chosen model.

By entering the values of xx and ln⁡p\ln p into a graphing calculator and performing a linear regression of ln⁡p\ln p on xx:
a=−12.465≈−12.5a = -12.465 \approx -12.5 (to 3 s.f.)
b=0.19783≈0.198b = 0.19783 \approx 0.198 (to 3 s.f.)

Answer

a=−12.5a = -12.5, b=0.198b = 0.198

9(iv)

1. Substitute $x = 19$ into the regression model for C to find the estimated probability.

From part (iii), the constants for Model C (ln⁡p=a+bx\ln p = a + bx) are a≈−12.465a \approx -12.465 and b≈0.19783b \approx 0.19783.
Substituting x=19x = 19,
ln⁡p=−12.46508+0.1978308(19)\ln p = -12.46508 + 0.1978308(19)
ln⁡p=−8.7063\ln p = -8.7063
p=e−8.7063≈0.000166 (to 3 s.f.)p = e^{-8.7063} \approx 0.000166 \text{ (to 3 s.f.)}

2. Comment on the reliability of the estimate.

The estimate is unreliable because x=19x = 19 is outside the range of given data values (25≤x≤4525 \le x \le 45), meaning this is an extrapolation.

Answer

0.0001660.000166

The estimate is unreliable as x=19x=19 lies outside the given data range (extrapolation).

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