Math Problem Statement

A simple random sample of size nequals15 is drawn from a population that is normally distributed. The sample mean is found to be x overbarequals22.2 and the sample standard deviation is found to be sequals6.3. Determine if the population mean is different from 25 at the alpha equals 0.01 level of significance. Complete parts ​(a) through ​(d) below. Question content area bottom Part 1 ​(a) Determine the null and alternative hypotheses. Upper H 0​: mu equals 25 Upper H 1​: mu not equals 25 Part 2 ​(b) Calculate the​ P-value. ​P-valueequals

0.107 ​(Round to three decimal places as​ needed.) Part 3 ​(c) State the conclusion for the test. A. Reject Upper H 0 because the​ P-value is greater than the alphaequals0.01 level of significance. B. Do not reject Upper H 0 because the​ P-value is less than the alphaequals0.01 level of significance. C. Do not reject Upper H 0 because the​ P-value is greater than the alphaequals0.01 level of significance. D. Reject Upper H 0 because the​ P-value is less than the alphaequals0.01 level of significance.

Solution

To address the problem, we'll follow the standard procedure for hypothesis testing. Let's go through each part step by step.

Part 1: State the Hypotheses

The null hypothesis (H₀) and the alternative hypothesis (H₁) are given as:

  • Null Hypothesis (H₀): μ=25\mu = 25
  • Alternative Hypothesis (H₁): μ25\mu \neq 25

Part 2: Calculate the P-value

We need to perform a t-test to determine if the sample mean is significantly different from the population mean. The formula for the test statistic (t) is:

t=xˉμs/nt = \frac{\bar{x} - \mu}{s / \sqrt{n}}

Where:

  • xˉ=22.2\bar{x} = 22.2 (sample mean)
  • μ=25\mu = 25 (population mean)
  • s=6.3s = 6.3 (sample standard deviation)
  • n=15n = 15 (sample size)

First, we calculate the test statistic tt:

t=22.2256.3/15t = \frac{22.2 - 25}{6.3 / \sqrt{15}}

Calculate the denominator:

6.315=6.33.872981.6266\frac{6.3}{\sqrt{15}} = \frac{6.3}{3.87298} \approx 1.6266

Then the test statistic:

t=22.2251.62662.81.62661.721t = \frac{22.2 - 25}{1.6266} \approx \frac{-2.8}{1.6266} \approx -1.721

Next, we find the degrees of freedom (df):

df=n1=151=14df = n - 1 = 15 - 1 = 14

Using a t-distribution table or a calculator for a two-tailed test, we find the p-value for t=1.721t = -1.721 with 14 degrees of freedom.

According to the information given in the question, the p-value is approximately 0.107.

Part 3: State the Conclusion

We compare the p-value with the significance level (α\alpha) to make a decision. Given:

  • α=0.01\alpha = 0.01
  • p-value=0.107\text{p-value} = 0.107

Since the p-value (0.107) is greater than the significance level (0.01), we do not reject the null hypothesis.

Therefore, the correct conclusion is:

C. Do not reject H0H_0 because the p-value is greater than the α=0.01\alpha = 0.01 level of significance.

Summary

  • Hypotheses: H0:μ=25H_0: \mu = 25, H1:μ25H_1: \mu \neq 25
  • P-value: 0.107
  • Conclusion: Do not reject H0H_0

Would you like more details or have any questions?

Here are five questions you can ask next:

  1. How is the t-distribution different from the normal distribution?
  2. What are the assumptions of the t-test?
  3. How can we interpret the p-value in hypothesis testing?
  4. Can we use this method if the population is not normally distributed?
  5. How does sample size affect the t-test?

Tip: When conducting hypothesis tests, always ensure your sample size and data meet the necessary assumptions to validate the results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Normal Distribution
T-Distribution
P-Value

Formulas

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

Theorems

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Suitable Grade Level

Undergraduate Level