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Statistics and Probability - Hypothesis testing (HL)

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Null Hypothesis (H0H_0): The statement that there is no effect, no difference, or no relationship between variables. It is assumed true until evidence suggests otherwise.

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Alternative Hypothesis (H1H_1): The claim that there is a significant effect, difference, or relationship. This can be one-tailed (e.g., μ>μ0\mu > \mu_0) or two-tailed (e.g., μ≠μ0\mu \neq \mu_0).

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Type I Error: Occurs when H0H_0 is rejected but is actually true. The probability of a Type I error is equal to the significance level α\alpha.

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Type II Error: Occurs when H0H_0 is not rejected but is actually false (i.e., H1H_1 is true). The probability is denoted by β\beta.

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P-value: The probability of obtaining the observed results (or more extreme) assuming H0H_0 is true. If p<αp < \alpha, we reject H0H_0.

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χ2\chi^2 Test for Independence: Used to determine if two categorical variables are independent. Degrees of freedom df=(r−1)(c−1)df = (r-1)(c-1).

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χ2\chi^2 Goodness of Fit Test: Used to check if observed data fits a specific theoretical distribution (Uniform, Binomial, Poisson, or Normal).

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One-sample tt-test: Compares the mean of a single sample to a known population mean μ\mu when the population standard deviation σ\sigma is unknown.

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Two-sample tt-test: Compares the means of two independent groups to see if they come from populations with the same mean (μ1=μ2\mu_1 = \mu_2).

📐Formulae

χcalc2=∑i=1k(fo−fe)2fe\chi^2_{calc} = \sum_{i=1}^{k} \frac{(f_o - f_e)^2}{f_e}

fe=row total×column totalgrand totalf_e = \frac{\text{row total} \times \text{column total}}{\text{grand total}}

t=xˉ−μsn−1nt = \frac{\bar{x} - \mu}{\frac{s_{n-1}}{\sqrt{n}}}

t=dˉ−μdsdnt = \frac{\bar{d} - \mu_d}{\frac{s_d}{\sqrt{n}}} (for paired tt-tests)

df=(r−1)(c−1)df = (r-1)(c-1) (for independence tests)

df=k−1−mdf = k - 1 - m (for goodness of fit, where mm is the number of estimated parameters)

💡Examples

Problem 1:

A school wants to test if a new teaching method improves math scores. A group of 88 students is tested before and after the intervention. The differences (After −- Before) are: 5,3,−1,4,2,6,0,45, 3, -1, 4, 2, 6, 0, 4. Perform a paired tt-test at the 5%5\% significance level to see if scores improved.

Solution:

H0:μd=0H_0: \mu_d = 0 (no improvement), H1:μd>0H_1: \mu_d > 0 (improvement). Using the data, the mean difference is dˉ=2.875\bar{d} = 2.875 and the sample standard deviation is sd≈2.416s_d \approx 2.416. n=8n = 8. The tt-statistic is t=2.875−02.416/8≈3.366t = \frac{2.875 - 0}{2.416 / \sqrt{8}} \approx 3.366. Using a GDC, the pp-value for a one-tailed test with df=7df = 7 is p≈0.006p \approx 0.006.

Explanation:

Because p=0.006<0.05p = 0.006 < 0.05, we reject H0H_0. There is significant evidence at the 5%5\% level to suggest the new teaching method improves scores.

Problem 2:

In a χ2\chi^2 test for independence between 'Gender' (2 categories) and 'Subject Preference' (3 categories: Math, Science, Art), the total sample size is 200200. Calculate the degrees of freedom and the expected frequency for 'Males' who prefer 'Math' if there are 9090 males in total and 7070 students in total prefer Math.

Solution:

df=(r−1)(c−1)=(2−1)(3−1)=1×2=2df = (r-1)(c-1) = (2-1)(3-1) = 1 \times 2 = 2 Expected frequency fef_e for (Male, Math): fe=90×70200=6300200=31.5f_e = \frac{90 \times 70}{200} = \frac{6300}{200} = 31.5

Explanation:

Degrees of freedom for a contingency table are calculated based on the number of rows and columns. The expected frequency assumes the null hypothesis (independence) is true.

Problem 3:

A manufacturer claims that the average lifespan of a battery is 5050 hours. A random sample of 1515 batteries is tested, yielding a mean of 48.248.2 hours and a standard deviation of 2.52.5 hours. Test this claim against the alternative that the lifespan is less than 5050 hours at α=0.01\alpha = 0.01.

Solution:

H0:μ=50H_0: \mu = 50, H1:μ<50H_1: \mu < 50. Using a one-sample tt-test: xˉ=48.2\bar{x} = 48.2, s=2.5s = 2.5, n=15n = 15. t=48.2−502.5/15≈−2.7885t = \frac{48.2 - 50}{2.5/\sqrt{15}} \approx -2.7885 Using GDC with df=14df = 14, pp-value ≈0.0073\approx 0.0073.

Explanation:

Since p=0.0073<0.01p = 0.0073 < 0.01, we reject H0H_0. There is sufficient evidence to conclude that the mean lifespan is significantly less than 5050 hours.