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Georgianna claims that in a small city renowned for its music school, the average child takes at least 5 years of piano lessons. We have a random sample of 30 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Use a hypothesis test to determine if there is sufficient evidence against Georgianna's claim. (b) Construct a \(95 \%\) confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. (c) Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.

Short Answer

Expert verified
We fail to reject Georgianna's claim; the evidence isn't sufficient to claim the mean is less than 5 years. The confidence interval includes 5, aligning with this outcome.

Step by step solution

01

Define the Hypotheses

First, we need to define the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the average number of years children take piano lessons is at least 5 years, or \( \mu \geq 5 \). The alternative hypothesis \( H_a \) is that the average number of years is less than 5, or \( \mu < 5 \).
02

Calculate the Test Statistic

To calculate the test statistic for a t-test, we use the formula \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \( \bar{x} = 4.6 \) is the sample mean, \( \mu = 5 \) is the hypothesized population mean, \( s = 2.2 \) is the sample standard deviation, and \( n = 30 \) is the sample size.\[ t = \frac{4.6 - 5}{2.2 / \sqrt{30}} = \frac{-0.4}{0.401} \approx -0.9975 \]
03

Determine the Critical Value and Decision Rule

For a one-sample t-test with 29 degrees of freedom at a 0.05 significance level, find the critical t-value for a one-tailed test. Using a t-distribution table or calculator, the critical value is approximately -1.699. If the calculated t-value is less than the critical value, we reject the null hypothesis.
04

Make a Decision

The calculated t-value is \(-0.9975\), which is greater than the critical value of \(-1.699\). Therefore, we fail to reject the null hypothesis. There is not enough evidence to claim that the average number of years is less than 5.
05

Construct the Confidence Interval

To construct a 95% confidence interval for the mean, use the formula \[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \] where \( \bar{x} = 4.6 \), \( s = 2.2 \), \( n = 30 \), and \( t^* \) is the critical value for a 95% confidence level which is 2.045.This gives us:\[ 4.6 \pm 2.045 \times \frac{2.2}{\sqrt{30}} = 4.6 \pm 0.821 \]Thus, the interval is \([3.779, 5.421]\).
06

Interpret the Confidence Interval

The 95% confidence interval suggests that we are 95% confident that the true mean number of years children in the city take piano lessons is between 3.779 and 5.421 years. Since the interval includes 5, it aligns with failing to reject the null hypothesis.
07

Compare Test and Confidence Interval Results

Both the hypothesis test and the confidence interval suggest there is not enough evidence to reject the claim that children, on average, take at least 5 years of piano lessons. The confidence interval includes the value 5, indicating it's a plausible value for the population mean.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval provides a range of values that likely contain the true population parameter, such as the mean. In our example, we are focusing on the average number of years children take piano lessons in the city. To calculate a confidence interval, we use the sample data and apply a formula that incorporates a critical value from a distribution, which in this case is the t-distribution.

The formula for the confidence interval is:
  • Sample mean (\(\bar{x}\)) plus or minus the critical value (\(t^*\)) times the standard error.
  • Standard error is calculated as the sample standard deviation (\(s\)) divided by the square root of the sample size (\(n\)).
This results in a range in which we believe the true population mean will fall 95% of the time if we were to repeat the sampling process. In this exercise, our interval is \([3.779, 5.421]\), which suggests the average number of years of piano lessons lies in this range.
T-Distribution
The t-distribution is a crucial concept in hypothesis testing, particularly when dealing with small sample sizes or when the population standard deviation is unknown. It helps us to approximate the sample means and adjust for uncertainty. The t-distribution resembles a normal distribution but has heavier tails, which means there is more probability in the extremes.

When conducting a t-test, as in our exercise, the degrees of freedom play a significant role. They are typically calculated as the sample size minus one (\(n - 1\)). In our case, with a sample of 30 children, the degrees of freedom are 29.
  • The shape of the t-distribution is determined by the degrees of freedom; higher degrees of freedom yield a distribution that more closely resembles the normal distribution.
  • We use a t-distribution table or calculator to find critical values necessary for hypothesis testing and confidence intervals.
Null Hypothesis
The null hypothesis (\(H_0\)) is a central building block in hypothesis testing. It represents a statement of no effect or no difference and is the hypothesis we aim to test against. In our piano lessons example, the null hypothesis posits that the average number of years children take lessons is at least 5 (\(\mu \geq 5\)). This acts as our default assumption that there is no significant deviation from this claim.

During hypothesis testing, our job is to either reject or fail to reject the null hypothesis. Failing to reject it means there isn't convincing evidence against it given the data we have.
  • This hypothesis helps set up the framework for testing, and its complement, the alternative hypothesis, provides a contrasting statement that can be supported by evidence.
  • Decisions are made based on the comparison of the calculated statistic and critical values tied to the significance level (e.g., 0.05).
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) provides a statement that we test against the null hypothesis. It claims there is a significant effect or difference. For example, in the context of music lessons in the city, the alternative hypothesis suggests that the average number of years children take piano lessons is less than 5 (\(\mu < 5\)).

This hypothesis is what researchers often hope to find evidence to support, showing that the default assumption (null hypothesis) is unlikely given the data.
  • When we perform tests, such as the t-test in our exercise, a result leads us to either support the alternative hypothesis or stick with the null based on the data's outcomes.
  • It is essential because it directs the inferential analysis by outlining the expected deviation from the claim outlined in the null hypothesis.

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Most popular questions from this chapter

The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. (a) Raina wants to use a \(90 \%\) confidence interval. How large a sample should she collect? (b) Luke wants to use a \(99 \%\) confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina's, and explain your reasoning. (c) Calculate the minimum required sample size for Luke.

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A \(90 \%\) confidence interval for a population mean is \((65,77) .\) The population distribution is approximately normal and the population standard deviation is unknown. This confidence interval is based on a simple random sample of 25 observations. Calculate the sample mean, the margin of error, and the sample standard deviation.

Air quality measurements were collected in a random sample of 25 country capitals in 2013, and then again in the same cities in 2014. We would like to use these data to compare average air quality between the two years. Should we use a paired or non-paired test? Explain your reasoning.

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