A survey on remote learning was carried out on a random sample of 400 students - Leaving Cert Mathematics - Question 5 - 2022

First calculate the margin of error using:

$ME = 1.96 \times \sqrt{\frac{p(1-p)}{n}}$

Substituting the known values:

• $p = 0.3375$
• $n = 400$

Calculate: $ME = 1.96 \times \sqrt{\frac{0.3375 \times (1 - 0.3375)}{400}}$

Calculating the confidence interval:

$CI = p \pm ME = 0.3375 \pm 0.05$

This leads to the interval: $[0.2875, 0.3875]$

Using:

$\hat{p} \pm 1.96 \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

Substituting:

• $\hat{p} = 0.3375$
• $n = 400$

We have: $CI = 0.3375 \pm 1.96 \times \sqrt{\frac{0.3375 \times (1 - 0.3375)}{400}}$

Performing the calculations yields: $[0.2912, 0.3838]$

The null hypothesis (H0):

• The average monthly spending for pre-pay plans remains at €20.79.

The alternative hypothesis (H1):

• The average monthly spending for pre-pay plans is different from €20.79.

To perform the hypothesis test:

• Use the formula for the z-score:

$z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$

Substituting in:

• $\bar{x} = 22.16$ (sample mean)
• $\mu = 20.79$ (population mean)
• $s = 8$ (standard deviation)
• $n = 500$ (sample size)

Calculating:

$z = \frac{22.16 - 20.79}{\frac{8}{\sqrt{500}}} \approx 3.7726$

After finding the z-value of approximately 3.7726, we compare it to the critical z-value for a 5% significance level (approximately 1.96). Since our calculated z-score exceeds 1.96, we reject the null hypothesis.

There is sufficient evidence at the 5% level of significance to conclude that the average monthly spending on mobile phones for people with a pre-pay plan has significantly changed from €20.79.