In a survey, 1000 people are selected at random and asked some questions about online shopping - Leaving Cert Mathematics - Question 3 - 2020

Let ar{p} = \frac{762}{1000} = 76.2\%. The margin of error from Part (i) is $3.2\%$.

The 95% confidence interval is calculated as:

$\left[ \bar{p} - ME, \bar{p} + ME \right]$

This gives us:

$\left[ 76.2\% - 3.2\%, 76.2\% + 3.2\% \right] = [73.0\%, 79.4\%]$

Thus, the 95% confidence interval is $[73.0\%, 79.4\%]$.

Let:

• Null hypothesis ($H_0$): $p = 0.80$
• Alternative hypothesis ($H_1$): $p \neq 0.80$

From the confidence interval calculated in Part (ii), we have: $[73.0\%, 79.4\%]$

Since $80\%$ is not within this confidence interval, we reject the null hypothesis.

Conclusion: There is sufficient evidence to reject $H_0$, concluding that the percentage of people who believe it is safe to give credit card details online is not $80\%$.

For a normal distribution with a mean of 480 and a standard deviation of 90, the values for $A$ and $B$ representing one standard deviation from the mean are:

• $A = 480 - 90 = 390$
• $B = 480 + 90 = 570$.

According to the Empirical Rule, approximately 68% of data falls within one standard deviation from the mean in a normal distribution.

Thus, for 6500 candidates:

$0.68 \times 6500 = 4420$

Therefore, approximately 4420 candidates are within the shaded region.