Lecture 19

HypothesisHypothesisTestingTestingforforMeansMeans--LargeLargeSamplesSamples

Remark

Hypothesis testing for a population mean is a statistical method used to evaluate claims about the average value of a population. It involves setting up two competing hypotheses: the null hypothesis $H_0$, which states the population mean is equal to a specific value $\mu_0$, and the alternative hypothesis $H_1$, which proposes the mean differs from $\mu_0$ (can be greater than, less than, or not equal to). We then collect data from a sample of the population and calculate a test statistic to assess how likely it is to observe such a sample mean if the null hypothesis were true. Finally, based on a pre-defined significance level $\alpha$ (usually 0.05), we decide whether to reject the null hypothesis. Rejecting $H_0$ suggests the sample data provides evidence against the assumed population mean, lending weight to the alternative hypothesis.

Remark

We will start by discussing hypothesis testing for population means when the sample size is large ($n > 30$). In this case, the sampling distribution of the sample mean is approximately normally distributed, and the test will be based on the standard normal distribution ($z$-score). The three types of hypothesis tests for population means can be represented pictorially. The null hypothesis is rejected if the test statistic falls within the rejection region.

Right-tailed test: $H_0: \mu = \mu_0$ vs. $H_1: \mu > \mu_0$

Left-tailed test: $H_0: \mu = \mu_0$ vs. $H_1: \mu < \mu_0$

Two-tailed test: $H_0: \mu = \mu_0$ vs. $H_1: \mu \neq \mu_0$

Example 1

The lifespan of honey badgers in captivity is on average $24$ years. The average lifespan of honey badgers in the wild is unknown. A large long-running study tracking $52$ honey badgers in the wild found that the sample average lifespan was $21.7$ years with a sample standard deviation of $7.3$ years. Let $\mu$ be the average lifespan of honey badgers in the wild. Test $H_0: \mu = 24$ against $H_1: \mu < 24$ at the $\alpha = 0.05$ significance level.

The test statistic is given by:$$z = \frac{\overline{X} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{21.7 - 24}{\frac{7.3}{\sqrt{52}}} = -2.27 $$The $p$-value is $P(z < -2.27) = 0.0116$. Since $0.0116 < 0.05$, we reject the null hypothesis. There is sufficient evidence to suggest that the average lifespan of honey badgers in the wild is less than $24$ years.

Example 2

The average weight of female American badgers is $6.7$ kg. It is conjectured that the Northern populations of American badgers are heavier. A sample of $36$ female Northern American badgers was taken in Alberta, and the average weight was found to be $9.5$ kg with a standard deviation of $2.2$ kg. Test $H_0: \mu = 6.7$ against $H_1: \mu > 6.7$ at the $\alpha = 0.01$ significance level.

The test statistic is given by:$$z = \frac{\overline{X} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{9.5 - 6.7}{\frac{2.2}{\sqrt{36}}} = 7.64 $$The $p$-value is $P(z > 7.64) \approx 0$. Since $0 < 0.01$, we reject the null hypothesis. There is sufficient evidence to suggest that the average weight of Northern female badgers is greater than the American average.

Example 3

It is rumored that the bite force of the spotted hyena is $1100$ psi. (Your tires are at $32-35$ psi.) To check the claim a team of rangers and biologists took a sample of $118$ hyenas and measured their bite force. The sample average was $1064$ psi with a sample standard deviation of $260$ psi. Test $H_0: \mu = 1100$ against $H_1: \mu < 1100$ at the $\alpha = 0.05$ significance level.

The test statistic is given by:$$z = \frac{\overline{X} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{1064 - 1100}{\frac{260}{\sqrt{118}}} = -1.47 $$The $p$-value is $P(z < -1.47) = 0.0708$. Since $0.0708 > 0.05$, we fail to reject the null hypothesis. There is insufficient evidence to suggest that the bite force of the spotted hyena is less than $1100$ psi.

Lecture 18 Lecture 20