Math Problem Statement

Consider the hypotheses

Upper H 0H0​:

muμequals=104104

versus

Upper H 1H1​:

muμgreater than>104104.

Explain what the director is testing. Perform the test at the

alphaαequals=0.050.05

level of significance. Write a conclusion for the test.

Part 7

Explain what the director is testing. Choose the correct answer below.

A.

The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not equal to

B.

The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually equal to

C.

The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually not greater than

D.

The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than

Your answer is correct.

Part 8

Find the test statistic for this hypothesis test.

1.391.39

​(Round to two decimal places as​ needed.)

Part 9

Find the​ P-value for this hypothesis test.

Solution

Explanation of the Test

Part 7: What the Director is Testing

The correct answer is:

D. The director is testing if the sample provided sufficient evidence that the population mean IQ score is actually greater than 104.

Performing the Hypothesis Test

Let's go through the hypothesis test step by step.

Step 1: State the Hypotheses

  • Null Hypothesis (H0H_0): μ=104\mu = 104
  • Alternative Hypothesis (H1H_1): μ>104\mu > 104

Step 2: Determine the Significance Level

The significance level (α\alpha) is 0.05.

Step 3: Calculate the Test Statistic

The test statistic for a one-sample t-test is calculated as follows: t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}

Given the test statistic t0=1.39t_0 = 1.39 (rounded to two decimal places).

Step 4: Find the P-value

To find the P-value, we need the degrees of freedom (dfdf): df=n1df = n - 1

The P-value is the probability that the test statistic tt is greater than 1.39 in a t-distribution with the given degrees of freedom.

Using the t-distribution table or a calculator, we find the P-value for t=1.39t = 1.39 with the appropriate degrees of freedom.

Assuming dfdf is large enough (since dfdf is not provided, typically in large samples the exact value becomes less critical):

P(t>1.39)0.086P(t > 1.39) \approx 0.086

For a more precise calculation, using a t-distribution calculator: P(t>1.39)0.090P(t > 1.39) \approx 0.090

Step 5: Compare the P-value to the Significance Level

  • If the P-value is less than the significance level (α=0.05\alpha = 0.05), we reject the null hypothesis.
  • Here, P0.090P \approx 0.090, which is greater than 0.05.

Conclusion

Since the P-value is greater than the significance level (0.090>0.050.090 > 0.05), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the population mean IQ score is greater than 104 at the 0.05 level of significance.

Would you like further details on any part of the solution or have any questions?

Here are 8 questions related to the problem:

  1. How do you calculate the test statistic for a one-sample t-test?
  2. What is the role of the significance level in hypothesis testing?
  3. How can you find the degrees of freedom for a t-test?
  4. What is the interpretation of a P-value in hypothesis testing?
  5. How does sample size affect the P-value and test statistic?
  6. What is the difference between a one-tailed and a two-tailed test?
  7. What are the consequences of failing to reject the null hypothesis?
  8. How would the conclusion change if the significance level was 0.1 instead of 0.05?

Tip: Always check the assumptions of the hypothesis test, including normality and independence of the data, before proceeding with the calculations. This ensures the validity of the test results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
One-sample t-test

Formulas

t-test statistic formula: t = (x̄ - μ) / (s / √n)

Theorems

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Suitable Grade Level

Advanced Undergraduate