It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients. After some new initiatives were introduced, a random sample of 12 patients from the hospital's A&E Department had the following waiting times, in hours. 4.25 3.90 4.15 3.95 4.20 4.15 5.00 3.85 4.25 3.80 3.95 4.15 Carry out a hypothesis test at the 10% significance level to investigate whether the mean waiting time at this hospital's A&E department has changed. You may assume that the waiting times are normally distributed with standard deviation 0.8 hours.

A-Level AQA Maths Pure Modelling involving Numerical Methods: It is known that a hospital has

It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 3

Question 14

It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients. After some new initiatives were introduced, a random s... show full transcript

Worked Solution & Example Answer:It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 3

Step 1

State both hypotheses correctly for two-tailed test

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Answer

Let the population mean be denoted as $u$ . The null hypothesis ( $H_0$ ) is that the mean waiting time is 4 hours:

u = 4$$ The alternative hypothesis ($H_1$) is that the mean waiting time has changed: $$H_1: u eq 4$$

Step 2

Calculate means of the sample

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Answer

The sample waiting times are:

$X = [4.25, 3.90, 4.15, 3.95, 4.20, 4.15, 5.00, 3.85, 4.25, 3.80, 3.95, 4.15]$

To find the sample mean ( $ar{x}$ ):

$ar{x} = rac{4.25 + 3.90 + 4.15 + 3.95 + 4.20 + 4.15 + 5.00 + 3.85 + 4.25 + 3.80 + 3.95 + 4.15}{12} = 4.13$

Step 3

Formulate the test statistic

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Answer

Using the formula for the test statistic:

u}{ rac{ ext{SD}}{ ext{sqrt}(n)}}$$ where: - $ar{x}$ is the sample mean, - $ u$ is the population mean (4), - SD is the standard deviation (0.8), and - $n$ is the sample size (12). Substituting the values: $$T = rac{4.13 - 4}{ rac{0.8}{ ext{sqrt}(12)}} = rac{0.13}{0.2308} \\ herefore T ightarrow 0.563$$

Step 4

Determine the critical value and acceptance region

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Answer

For a two-tailed test at the 10% significance level, the critical values can be found from the t-distribution table for $df = n - 1 = 11$ :

The critical t-values at $rac{ ext{alpha}}{2} = 0.05$ are approximately $ext{t}_{0.05}(11) = ext{±}1.65$ .

Thus, the acceptance region is:

$-1.65 < T < 1.65$

Step 5

Compare the test statistic to critical values

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Our calculated test statistic ( $0.563$ ) lies within the acceptance region $(-1.65 < 0.563 < 1.65)$

Thus, we do not reject the null hypothesis ( $H_0$ ).

Step 6

Conclude correctly in context

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Answer

There is insufficient evidence to suggest that the mean waiting time at this hospital's A&E department has changed.

It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 3

Worked Solution & Example Answer:It is known that a hospital has a mean waiting time of 4 hours for its Accident and Emergency (A&E) patients - AQA - A-Level Maths Pure - Question 14 - 2020 - Paper 3

State both hypotheses correctly for two-tailed test

Calculate means of the sample

Formulate the test statistic

Determine the critical value and acceptance region

Compare the test statistic to critical values

Conclude correctly in context

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