Teams ? Details Objective: Use StatCrunch to find test statistics and P-values from a set of data to perform a formal hypothesis test. Assignment: Open the handout for directions to obtain the main data set from StatCrunch database. Once you complete have the main set in ascending order, you are to take your sample. Your sample data set is: The heights of children from Australia. Complete the following exercises listed below with that sample. Directions For data 1 sample-1.pdf 5 rary Resources m als 1. 5pts. Use a level of significance of 0.01 to test the claim that the population mean height of the children in Australia exceeds 150 cm. Provide the null and alternative hypotheses, test statistic (3 decimal places), the P-value (4 decimal places), the conclusions about the null hypothesis, the final conclusion. 2. 5pts. Use a level of significance of 0.05 to test the claim that the population variation in the height of the children in Australia is less than 20cm. Provide the null and alternative hypotheses, test statistic (3 decimal places), the P-value (4 decimal places), the conclusions about the null hypothesis, the final conclusion.
The Correct Answer and Explanation is:
Problem 1: Testing if the population mean height of Australian children exceeds 150 cm at a significance level of 0.01.
1. Null and Alternative Hypotheses:
- Null Hypothesis (H₀): The population mean height is less than or equal to 150 cm. H0:μ≤150H₀: \mu \leq 150H0:μ≤150
- Alternative Hypothesis (H₁): The population mean height is greater than 150 cm. H1:μ>150H₁: \mu > 150H1:μ>150
2. Test Statistic:
You would need to use a t-test for a single sample since you are testing the mean based on sample data. The formula for the test statistic is:t=xˉ−μ0snt = \frac{\bar{x} – \mu_0}{\frac{s}{\sqrt{n}}}t=nsxˉ−μ0
Where:
- xˉ\bar{x}xˉ = sample mean
- μ0\mu_0μ0 = hypothesized population mean (150 cm)
- sss = sample standard deviation
- nnn = sample size
You will need to input the heights of the children from Australia into StatCrunch to get the sample mean and sample standard deviation. After calculating the test statistic, compare it with the critical value or use the P-value to make the decision.
3. P-value:
The P-value will indicate how likely it is to observe the sample data if the null hypothesis were true. Use StatCrunch or a t-distribution table to calculate the P-value based on the test statistic.
4. Conclusion:
- Decision Rule: If the P-value is less than or equal to 0.01, reject the null hypothesis. If the P-value is greater than 0.01, fail to reject the null hypothesis.
- Final Conclusion: Based on the P-value, you will either reject the null hypothesis (supporting that the mean height exceeds 150 cm) or fail to reject it (indicating there isn’t enough evidence to support the claim).
Problem 2: Testing if the population variation in height is less than 20 cm at a significance level of 0.05.
1. Null and Alternative Hypotheses:
- Null Hypothesis (H₀): The population variance is greater than or equal to 20 cm². H0:σ2≥20H₀: \sigma^2 \geq 20H0:σ2≥20
- Alternative Hypothesis (H₁): The population variance is less than 20 cm². H1:σ2<20H₁: \sigma^2 < 20H1:σ2<20
2. Test Statistic:
This is a Chi-square test for variance. The formula for the test statistic is:χ2=(n−1)s2σ02\chi^2 = \frac{(n – 1) s^2}{\sigma^2_0}χ2=σ02(n−1)s2
Where:
- s2s^2s2 = sample variance
- σ02\sigma^2_0σ02 = hypothesized population variance (20 cm²)
- nnn = sample size
Use StatCrunch to calculate the sample variance from your data.
3. P-value:
The P-value for a Chi-square test will be calculated based on the test statistic. You can use StatCrunch to obtain this value and determine the significance of the test.
4. Conclusion:
- Decision Rule: If the P-value is less than or equal to 0.05, reject the null hypothesis. If the P-value is greater than 0.05, fail to reject the null hypothesis.
- Final Conclusion: Based on the P-value, decide whether the population variance is less than 20 cm² or not.
In both cases, once you calculate the test statistics and P-values using StatCrunch, you can make a decision based on the provided significance levels (0.01 and 0.05). If you need assistance with specific calculations or understanding the output from StatCrunch, let me know!
