In a region of England, the government decides to use an advertising campaign to encourage people to eat more healthily - AQA - A-Level Maths Mechanics - Question 18 - 2018 - Paper 3

This sample should not be used to conduct a hypothesis test because it is not randomly selected. The use of opportunistic sampling introduces potential biases, as those who are available may not represent the wider population. This lack of randomness violates one of the main assumptions required for valid hypothesis testing.

To investigate this, we set up the null and alternative hypotheses:

• Null hypothesis ( H_0): ext{Mean consumption} oldsymbol{eta} = 66.5g
• Alternative hypothesis ( H_1): ext{Mean consumption} oldsymbol{eta} < 66.5g

We calculate the test statistic using the formula:

z = rac{ar{x} - eta_0}{rac{ heta}{ ext{s.e.}}} Where:

• ar{x} = sample mean = 65.4g
• eta_0 = population mean = 66.5g
• heta = standard deviation = 21.2g
• s.e. = standard error = \frac{ heta}{\sqrt{n}} = \frac{21.2}{\sqrt{750}} \approx 0.73g

Calculating the value:

$z = \frac{65.4 - 66.5}{0.73} \approx -1.42$

Next, we state the critical z-value for a one-tailed test at the 10% level of significance. The critical z-value is approximately -1.28.

Since -1.42 < -1.28, we reject the null hypothesis, concluding that there is sufficient evidence to suggest that the advertising campaign has decreased the mean consumption of chocolate per person per week.