Final answer:
To test Georgianna's claim, the null hypothesis is μ = 5 and the alternative is μ > 5. After calculating the test statistic using the sample mean, standard deviation, and size, we then find the p-value and compare it to a significance level to make a conclusion.
Step-by-step explanation:
To evaluate Georgianna's claim that the average child in a small city takes more than 5 years of piano lessons, we should use a hypothesis test to compare the sample data to the claim. The correct null and alternative hypotheses to test whether the mean number of years, μ, is greater than 5 are:
- H0 (null hypothesis): μ = 5
- Ha (alternative hypothesis): μ > 5
Using the provided sample data with a mean of 5.4 years, a standard deviation of 2.2 years, and a sample size of 20, we can calculate the test statistic as follows:
T = (sample mean - population mean) / (standard deviation / sqrt(sample size))
T = (5.4 - 5) / (2.2 / sqrt(20))
T (rounded to two decimal places)
With the T calculated, we then compare it to the T-distribution with n-1 degrees of freedom to find the p-value.
The degrees of freedom (df) for this test are 20-1=19.
Once we have the p-value, we can compare it to the level of significance (usually 0.05 for a 5% significance level), to determine if we can reject the null hypothesis. If the p-value is less than the level of significance, we reject the null hypothesis, and the data provide strong evidence that the average number of years exceeds 5.
Without specific values for T or the p-value, it's not possible to definitively conclude here. If the calculated p-value is low, the conclusion is that the data do provide strong evidence that the average child in the city takes more than 5 years of piano lessons. However, if the p-value is high, then the data do not provide strong evidence for Georgianna's claim.