A pharmaceutical company claims that 60% of adults experience drowsiness as a side effect when taking their new allergy medication. A researcher suspects that the actual proportion is different from 60%. To investigate, the researcher randomly selects 225 adults who have taken the medication and finds that 120 of them experienced drowsiness.
Calculate an approximate 95% confidence interval for the proportion of adults who experience drowsiness when taking this medication. Based on this confidence interval, does the data support the pharmaceutical company’s claim? Explain your reasoning.
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Create Free Account Log inThis is a free VCE Units 3 & 4 Mathematical Methods practice question worth 5 marks, testing your understanding of Statistical Inference. It falls under Data analysis, probability and statistics in Unit 4: Mathematical Methods Unit 4. Submit your answer above to receive instant AI-powered marking and personalised feedback.
Continues the study of functions, algebra, calculus, and introduces probability and statistics.
Covers discrete and continuous random variables, probability distributions, and statistical inference for sample proportions.
statistical inference, including definition and distribution of sample proportions, simulations and confidence intervals: - distinction between a population parameter and a sample statistic and the use of the sample statistic to estimate the population parameter - simulation of random sampling, for a variety of values of $p$ and a range of sample sizes, to illustrate the distribution of $\hat{P}$ and variations in confidence intervals between samples - concept of the sample proportion $\hat{P}=\frac{x}{n}$ as a random variable whose value varies between samples, where $X$ is a binomial random variable which is associated with the number of items that have a particular characteristic and $n$ is the sample size - approximate normality of the distribution of $\hat{P}$ for large samples and, for such a situation, the mean $p$ (the population proportion) and standard deviation, $\sqrt{\frac{p(1-p)}{n}}$ - determination and interpretation of, from a large sample, an approximate confidence interval $\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)$, for a population proportion where $z$ is the appropriate quantile for the standard normal distribution, in particular the $95 \%$ confidence interval as an example of such an interval where $z \approx 1.96$ (the term standard error may be used but is not required).
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