A survey of 100 shoppers, randomly selected from a large number of Saturday supermarket shoppers, showed that the mean shopping spend was €90.45 - Leaving Cert Mathematics - Question 2 - 2015

Let’s set up the hypotheses:

• Null Hypothesis ($H_0$): The mean spend is €94, i.e., $\mu = 94$
• Alternative Hypothesis ($H_1$): The mean spend is not €94, i.e., $\mu \neq 94$

Calculate the z-score for the sample mean:

$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\text{sqrt}(n)}} = \frac{90.45 - 94}{\frac{20.73}{\text{sqrt}(100)}} = \frac{90.45 - 94}{2.073} = -1.71$

Using a z-table, we find two-tailed critical z-values for a 5% significance level are approximately ±1.96.

Conclusion: Since -1.71 is within the range of -1.96 and 1.96, we fail to reject the null hypothesis, indicating there is not enough evidence to reject the supermarket's claim.

To find the p-value related to the z-score of -1.71:

$P(z < -1.71) = 1 - P(z < 1.71) = 1 - 0.9564 = 0.0436$

Since this is a two-tailed test, the p-value is:

$p-value = 0.0436 \times 2 = 0.0872$

Meaning: This implies that if the mean amount spent really was €94, then the probability that the sample mean would be €90.45 by chance is 8.72%. Since this is more than a 5% chance, we do not reject the null hypothesis.