Normal Hypothesis Testing (Edexcel A-Level Mathematics): Revision Notes

Example: Testing the Mean Suppose a company claims that their light bulbs last an average of 1,000 hours. A consumer group believes that the true mean lifetime is less than $1,000$ $hours$ and wants to test this claim. They take a random sample of 50 bulbs and find an average lifetime of 980 hours with a standard deviation of 40 hours. They wish to test this at the 5% significance level.

Step 1: State the Hypotheses

• Null Hypothesis ($H_0$): $\mu = 1000$ (The mean lifetime is $1,000$ $hours$.)
• Alternative Hypothesis ($H_1$): $\mu < 1000$ (The mean lifetime is less than $1,000$ $hours$.)

Step 2: Choose the Significance Level

• Significance level ($\alpha$): 0.05

Step 3: Calculate the Test Statistic Since the sample size is large, the sampling distribution of the mean can be approximated by a Normal distribution. The test statistic $(z)$ is calculated as:

Where:

• $\bar{x} = 980$ (sample mean)
• $\mu_0 = 1000$ (hypothesised population mean)
• $\sigma = 40$ (sample standard deviation)
• $n = 50$ (sample size) Substitute the values:

Step 4: Find the Critical Value or P-Value For a significance level of $0.05$ in a one-tailed test, the critical $z-value$ is approximately -1.645.

Alternatively, you can find the $p-value$ associated with the test statistic:

• Using the $z-table$, $P(Z < -3.54)$ is very small, approximately 0.0002.

Step 5: Make a Decision

• Since the test statistic $z = -3.54$ is less than the critical value of -1.645, we reject the null hypothesis.
• Alternatively, because the $p-value$ (0.0002) is less than the significance level (0.05), we reject the null hypothesis. Conclusion: There is strong evidence at the $5$% significance level to suggest that the mean lifetime of the light bulbs is less than $1,000$ $hours$.