Understand and use the approximate margin of error. 𝐸 = 𝑧 𝑠 √𝑛

A market research organisation conducts a survey of household spending on groceries. A random sample of 150 households is selected, and the sample mean monthly spending is calculated to be $485 with a sample standard deviation of $67. Calculate the approximate margin of error at the 95% confidence level.

A researcher surveys a random sample of 225 households to estimate the average weekly household expenditure on groceries. The sample standard deviation is $42.50, and the researcher wants to construct a 95% confidence interval. (a) Calculate the margin of error using \(E = z \frac{s}{\sqrt{n}}\), where the critical value for a 95% confidence level is \(z = 1.96\). (1 mark) (b) If the sample mean is $156.80, determine the lower and upper bounds of the 95% confidence interval. (2 marks)

A market research firm surveyed a random sample of 150 consumers about their spending on streaming services. The sample yielded a mean expenditure of $18.50 per month with a sample standard deviation of $4.20. (a) Calculate the approximate margin of error at a 95% confidence level. (2 marks) (b) Interpret what this margin of error tells us about the estimate of the population mean monthly expenditure. (1 mark)

A researcher wishes to estimate the mean weekly screen time (hours) of university students. A random sample of 36 students yields a sample standard deviation of 4.2 hours. Using a 95% confidence level, what is the approximate margin of error for the sample mean?

A researcher conducts a survey on the time spent on social media by secondary school students. A random sample of 64 students is taken and the sample standard deviation is found to be 18 minutes. For a 95% confidence level, calculate the approximate margin of error. Round your answer to the nearest minute.