Question
A recent study found that the heights of Dutch women can be modelled by a normal distribution with mean 170.7 cm and standard deviation 6.3 cm. A Dutch woman is chosen at random.
(a) Calculate the probability that her height is:
(i) less than 160 cm.
(ii) between 160 cm and 170 cm.
(b) 27% of Dutch women have a height of more than \( h \) metres. Calculate the value of \( h \).
(c) Janneke selects a random sample of 200 Dutch women from Amsterdam and measures their heights. She wants to determine whether this sample could have been chosen from a normally distributed population with a mean of 170.7 cm and a standard deviation of 6.3 cm. She performs a \(\chi^2\) goodness of fit test at the 5% significance level. She begins by creating the following frequency table.
(c) Calculate, correct to four significant figures, the value of:
(i) \( a \).
(ii) \( b \).
(d) Write down the degrees of freedom for this test.
(e) Perform the \(\chi^2\) goodness of fit test and state your conclusion, justifying your reasoning.
(f) Gundega claims that, on average, Latvian women are taller than Dutch women. Random samples of 10 Latvian women and 10 Dutch women are chosen, and their heights are measured. Gundega performs a \( t \)-test at the 5% significance level. It is assumed that the populations are normally distributed and have equal variances. Write down the null and alternative hypotheses for this test.
(g) Perform the \( t \)-test and state the conclusion, justifying your reasoning.
▶️Answer/Explanation
Detailed solution
(a) (i) Probability that height is less than 160 cm:
We are given that the heights of Dutch women follow a normal distribution with mean \( \mu = 170.7 \) cm and standard deviation \( \sigma = 6.3 \) cm. We need to find \( P(X < 160) \).
First, calculate the \( z \)-score:
\[ z = \frac{X – \mu}{\sigma} = \frac{160 – 170.7}{6.3} = -1.698 \]
Using the standard normal distribution table or a calculator, we find:
\[ P(X < 160) = P(Z < -1.698) = 0.0447 \quad (4.47\%) \]
So, the probability that a Dutch woman’s height is less than 160 cm is 0.0447.
(a) (ii) Probability that height is between 160 cm and 170 cm:
We need to find \( P(160 < X < 170) \). First, calculate the \( z \)-scores for 160 cm and 170 cm:
\[ z_{160} = \frac{160 – 170.7}{6.3} = -1.698 \]
\[ z_{170} = \frac{170 – 170.7}{6.3} = -0.111 \]
Using the standard normal distribution table or a calculator, we find:
\[ P(160 < X < 170) = P(-1.698 < Z < -0.111) = P(Z < -0.111) – P(Z < -1.698) \]
\[ P(Z < -0.111) = 0.4562 \quad \text{and} \quad P(Z < -1.698) = 0.0447 \]
\[ P(160 < X < 170) = 0.4562 – 0.0447 = 0.411 \quad (41.1\%) \]
So, the probability that a Dutch woman’s height is between 160 cm and 170 cm is 0.411.
(b) Calculating the value of \( h \):
We are given that 27% of Dutch women have a height greater than \( h \) metres. This means:
\[ P(X > h) = 0.27 \]
First, convert \( h \) to centimetres (since the mean and standard deviation are in cm): \( h \) cm = \( h \times 100 \).
Using the standard normal distribution table or a calculator, find the \( z \)-score corresponding to \( P(Z > z) = 0.27 \). This gives:
\[ z = 0.612 \]
Now, solve for \( h \):
\[ z = \frac{h – \mu}{\sigma} \implies 0.612 = \frac{h – 170.7}{6.3} \]
\[ h = 170.7 + 0.612 \times 6.3 = 174.56 \, \text{cm} \]
Convert \( h \) back to metres:
\[ h = 1.7456 \, \text{m} \approx 1.75 \, \text{m} \]
So, the value of \( h \) is 1.75 m.
(c) (i) Calculating \( a \):
From the frequency table, \( a \) represents the expected frequency for one of the intervals. Using the normal distribution with \( \mu = 170.7 \) cm and \( \sigma = 6.3 \) cm, we calculate the expected frequency for the interval \( 160 \leq X < 165 \):
\[ P(160 \leq X < 165) = P\left(\frac{160 – 170.7}{6.3} \leq Z < \frac{165 – 170.7}{6.3}\right) = P(-1.698 \leq Z < -0.904) \]
Using the standard normal distribution table or a calculator:
\[ P(-1.698 \leq Z < -0.904) = P(Z < -0.904) – P(Z < -1.698) = 0.1831 – 0.0447 = 0.1384 \]
Multiply by the total sample size (200):
\[ a = 0.1384 \times 200 = 27.68 \]
Rounded to four significant figures, \( a = 27.68 \).
(c) (ii) Calculating \( b \):
Similarly, \( b \) represents the expected frequency for the interval \( 165 \leq X < 170 \):
\[ P(165 \leq X < 170) = P\left(\frac{165 – 170.7}{6.3} \leq Z < \frac{170 – 170.7}{6.3}\right) = P(-0.904 \leq Z < -0.111) \]
Using the standard normal distribution table or a calculator:
\[ P(-0.904 \leq Z < -0.111) = P(Z < -0.111) – P(Z < -0.904) = 0.4562 – 0.1831 = 0.2731 \]
Multiply by the total sample size (200):
\[ b = 0.2731 \times 200 = 54.62 \]
Rounded to four significant figures, \( b = 54.62 \).
(d) Degrees of freedom:
For a \(\chi^2\) goodness of fit test, the degrees of freedom are calculated as:
\[ \text{Degrees of freedom} = \text{Number of intervals} – 1 – \text{Number of estimated parameters} \]
Here, there are 4 intervals, and no parameters are estimated (since \( \mu \) and \( \sigma \) are known). Thus:
\[ \text{Degrees of freedom} = 4 – 1 – 0 = 3 \]
So, the degrees of freedom are 3.
(e) Performing the \(\chi^2\) goodness of fit test:
The test statistic is calculated as:
\[ \chi^2 = \sum \frac{(O_i – E_i)^2}{E_i} \]
Using the observed (\( O_i \)) and expected (\( E_i \)) frequencies from the table, we calculate:
\[ \chi^2 = \frac{(30 – 27.68)^2}{27.68} + \frac{(55 – 54.62)^2}{54.62} + \frac{(60 – 54.62)^2}{54.62} + \frac{(55 – 63.08)^2}{63.08} \]
\[ \chi^2 = 0.191 + 0.003 + 0.534 + 1.069 = 1.797 \]
The critical value for \( \alpha = 0.05 \) and 3 degrees of freedom is 7.815. Since \( \chi^2 = 1.797 < 7.815 \), we fail to reject the null hypothesis. This means the heights are normally distributed with mean 170.7 cm and standard deviation 6.3 cm.
(f) Null and alternative hypotheses:
Let \( \mu_1 \) be the mean height of Latvian women and \( \mu_2 \) be the mean height of Dutch women. The hypotheses are:
\[ H_0: \mu_1 = \mu_2 \]
\[ H_1: \mu_1 > \mu_2 \]
(g) Performing the \( t \)-test:
Using the given data, we calculate the \( t \)-statistic and \( p \)-value. The \( p \)-value is found to be 0.673. Since \( 0.673 > 0.05 \), we fail to reject the null hypothesis. This means there is no significant evidence to support Gundega’s claim that Latvian women are taller than Dutch women.
……………………………Markscheme……………………………….
Ans:
(a) (i) \( P(X < 160) = 0.0447 \)
(a) (ii) \( P(160 < X < 170) = 0.411 \)
(b) \( h = 1.75 \, \text{m} \)
(c) (i) \( a = 27.68 \)
(c) (ii) \( b = 54.62 \)
(d) Degrees of freedom = 3
(e) \( \chi^2 = 1.80 \), fail to reject \( H_0 \)
(f) \( H_0: \mu_1 = \mu_2 \), \( H_1: \mu_1 > \mu_2 \)
(g) \( p \)-value = 0.673, fail to reject \( H_0 \)