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A survey during 2013 investigated mean expenditure on bread and on alcohol - AQA - A-Level Maths Pure - Question 14 - 2019 - Paper 3

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A survey during 2013 investigated mean expenditure on bread and on alcohol. The 2013 survey obtained information from 12 144 adults. The survey revealed that the m... show full transcript

Worked Solution & Example Answer:A survey during 2013 investigated mean expenditure on bread and on alcohol - AQA - A-Level Maths Pure - Question 14 - 2019 - Paper 3

Step 1

Carry out a hypothesis test, at the 5% significance level

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Answer

To carry out the hypothesis test, we start by stating the null and alternative hypotheses:

  • Null Hypothesis, H0:μ=123H_0: \mu = 123 (the mean expenditure per adult on bread has not changed)
  • Alternative Hypothesis, H1:μ123H_1: \mu \neq 123 (the mean expenditure per adult on bread has changed)

Next, we calculate the test statistic using the formula:

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

Where:

  • xˉ=127\bar{x} = 127 (sample mean)
  • μ0=123\mu_0 = 123 (hypothesized population mean)
  • s=70s = 70 (standard deviation)
  • n=12144n = 12144 (sample size)

Substituting the values into the formula gives:

t=12712370/12144=470/110.25=4×110.25706.30t = \frac{127 - 123}{70 / \sqrt{12144}} = \frac{4}{70 / 110.25} = \frac{4 \times 110.25}{70} \approx 6.30

Now, we find the critical values for a two-tailed test at the 5% significance level. Looking up values from the t-distribution table or using a calculator, we find that the critical values are approximately ±1.96.

Since the calculated test statistic (6.30) is greater than 1.96, we reject the null hypothesis. This indicates that there is evidence to suggest that the mean expenditure per adult per week on bread changed from 2012 to 2013.

Step 2

Calculate the greatest and least values for the sample mean expenditure

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Answer

To find the greatest and least sample means for which we would accept the null hypothesis, we use the critical value approach:

We find the acceptance region for the null hypothesis:

For a two-tailed test, the acceptance region is:

1.96<z<1.96-1.96 < z < 1.96

Calculating the corresponding upper and lower bounds:

  1. Upper Bound:

    127+1.96×(7012144)127 + 1.96 \times \left(\frac{70}{\sqrt{12144}}\right)

  2. Lower Bound:

    1271.96×(7012144)127 - 1.96 \times \left(\frac{70}{\sqrt{12144}}\right)

Calculating the values:

Using a calculator, we find: 70121440.63\frac{70}{\sqrt{12144}} \approx 0.63

Then:

  • Upper Bound: 127+1.96×0.63127+1.2358128.24127 + 1.96 \times 0.63 \approx 127 + 1.2358 \approx 128.24

  • Lower Bound: 1271.96×0.631271.2358125.76127 - 1.96 \times 0.63 \approx 127 - 1.2358 \approx 125.76

Thus, the greatest value is approximately 128.24 and the least value is approximately 125.76.

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