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Edexcel IAL - Statistics 3- 3.5 Hypothesis Tests for the Mean (Variance Known)- Study notes  - New syllabus

Edexcel IAL – Statistics 3- 3.5 Hypothesis Tests for the Mean (Variance Known) -Study notes- New syllabus

Edexcel IAL – Statistics 3- 3.5 Hypothesis Tests for the Mean (Variance Known) -Study notes -Edexcel A level Maths- per latest Syllabus.

Key Concepts:

  • 3.5 Hypothesis Tests for the Mean (Variance Known)

Edexcel IAL Maths-Study Notes- All Topics

Hypothesis Tests for the Mean of a Normal Distribution (Variance Known)

Hypothesis testing is a formal procedure used to decide whether there is sufficient evidence from sample data to support a claim about a population parameter. In this section, we consider tests for the population mean when the population variance is known and the population is normally distributed.

Assumptions

The following conditions apply:

  • The population distribution is normal
  • The population variance \( \mathrm{\sigma^2} \) is known
  • A random sample of size \( \mathrm{n} \) is taken

Null and Alternative Hypotheses

A hypothesis test begins with two competing hypotheses:

Null hypothesis \( \mathrm{H_0} \): the population mean has a specified value

Alternative hypothesis \( \mathrm{H_1} \): the population mean differs from the value in \( \mathrm{H_0} \)

Common forms of the alternative hypothesis are:

Two-tailed: \( \mathrm{H_1:\mu \neq \mu_0} \)

Upper-tailed: \( \mathrm{H_1:\mu > \mu_0} \)

Lower-tailed: \( \mathrm{H_1:\mu < \mu_0} \)

Test Statistic

The test statistic used is

\( \mathrm{Z = \dfrac{\bar{X} – \mu_0}{\sigma/\sqrt{n}}} \)

Under the null hypothesis, this statistic has the distribution

\( \mathrm{Z \sim N(0,1)} \)

Significance Level

The significance level, denoted by \( \mathrm{\alpha} \), is the probability of rejecting the null hypothesis when it is true.

Common significance levels are:

5% (\( \mathrm{\alpha = 0.05} \))

1% (\( \mathrm{\alpha = 0.01} \))

Critical Region Method

Using the critical region approach:

For a two-tailed test at 5% level: reject \( \mathrm{H_0} \) if \( \mathrm{|Z| > 1.96} \)

For an upper-tailed test at 5% level: reject \( \mathrm{H_0} \) if \( \mathrm{Z > 1.645} \)

For a lower-tailed test at 5% level: reject \( \mathrm{H_0} \) if \( \mathrm{Z < -1.645} \)

p-value Method

Alternatively, the p-value may be used.

The p-value is the probability of obtaining a test statistic at least as extreme as the observed value, assuming \( \mathrm{H_0} \) is true

Reject \( \mathrm{H_0} \) if the p-value is less than \( \mathrm{\alpha} \)

Link with Confidence Intervals

For a two-tailed test at significance level \( \mathrm{\alpha} \):

Reject \( \mathrm{H_0} \) if the hypothesised mean \( \mathrm{\mu_0} \) lies outside the \( \mathrm{(1-\alpha)\times100\%} \) confidence interval

Do not reject \( \mathrm{H_0} \) if it lies inside the interval

Key Points to Remember

  • The test statistic follows \( \mathrm{N(0,1)} \) under \( \mathrm{H_0} \)
  • Known variance allows direct use of the normal distribution
  • Decisions are based on critical values or p-values

Example :

The mean length of metal rods is claimed to be \( \mathrm{50\;cm} \). The population standard deviation is known to be \( \mathrm{4\;cm} \).

A random sample of \( \mathrm{n = 36} \) rods has a mean length of \( \mathrm{51.2\;cm} \).

Test the claim at the 5% significance level.

▶️ Answer/Explanation

Step 1: Hypotheses

\( \mathrm{H_0:\mu = 50} \)

\( \mathrm{H_1:\mu \neq 50} \)

Step 2: Test statistic

\( \mathrm{Z = \dfrac{51.2 – 50}{4/\sqrt{36}} = 1.8} \)

Step 3: Critical value

At 5% level, \( \mathrm{|Z| > 1.96} \) is the rejection region

Decision: \( \mathrm{1.8 < 1.96} \), so do not reject \( \mathrm{H_0} \).

Conclusion: There is insufficient evidence at the 5% level to suggest the mean length differs from 50 cm.

Example :

The mean time taken to complete a task is believed to be \( \mathrm{30\;minutes} \). The population standard deviation is \( \mathrm{6\;minutes} \).

A random sample of \( \mathrm{n = 25} \) tasks has a mean time of \( \mathrm{33\;minutes} \).

Test at the 5% significance level whether the mean time has increased.

▶️ Answer/Explanation

Step 1: Hypotheses

\( \mathrm{H_0:\mu = 30} \)

\( \mathrm{H_1:\mu > 30} \)

Step 2: Test statistic

\( \mathrm{Z = \dfrac{33 – 30}{6/\sqrt{25}} = 2.5} \)

Step 3: Critical value

Upper-tailed 5% test: reject if \( \mathrm{Z > 1.645} \)

Decision: \( \mathrm{2.5 > 1.645} \), so reject \( \mathrm{H_0} \).

Conclusion: There is sufficient evidence at the 5% level to conclude that the mean time has increased.

Example :

A manufacturer claims that the mean mass of packets is \( \mathrm{500\;g} \). The population standard deviation is \( \mathrm{5\;g} \).

A random sample of \( \mathrm{n = 100} \) packets has a mean mass of \( \mathrm{498.8\;g} \).

Test the claim at the 1% significance level.

▶️ Answer/Explanation

Step 1: Hypotheses

\( \mathrm{H_0:\mu = 500} \)

\( \mathrm{H_1:\mu \neq 500} \)

Step 2: Test statistic

\( \mathrm{Z = \dfrac{498.8 – 500}{5/\sqrt{100}} = -2.4} \)

Step 3: Critical value

At 1% level, reject if \( \mathrm{|Z| > 2.576} \)

Decision: \( \mathrm{|−2.4| < 2.576} \), so do not reject \( \mathrm{H_0} \).

Conclusion: There is insufficient evidence at the 1% level to conclude that the mean mass differs from 500 g.

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