Math Problem Statement

In conducting the hypothesis test below, your sample consists of 23 observations, with a mean of 57 and standard deviation of 3.44.

H 0 : μ ≤ 55.1

H 1 : μ

55.1

a) This is a right-tailed test. Correct

b) Calculate the test statistic, rounded to 3 decimal places. t

3.162 Incorrect

c) At α

0.1 , the rejection region is

t < -1.717 or t

1.717

t < -1.321

t

1.321

none of the above

Correct

d) The decision is to fail to reject H 0 since the test statistic does not fall in the rejection region. reject H 0 since the test statistic falls in the rejection region. fail to reject H 1 since the test statistic does not fall in the rejection region. reject H 1 since the test statistic falls in the rejection region. none of the above

Solution

To solve this hypothesis test, we need to follow these steps:

Given Data:

  • Sample size (nn) = 23
  • Sample mean (xˉ\bar{x}) = 57
  • Sample standard deviation (ss) = 3.44
  • Null hypothesis (H0H_0): μ55.1\mu \leq 55.1
  • Alternative hypothesis (H1H_1): μ>55.1\mu > 55.1
  • Significance level (α\alpha) = 0.1

a) This is a right-tailed test.

  • Correct: Since H1H_1 is μ>55.1\mu > 55.1, it is indeed a right-tailed test.

b) Calculate the test statistic:

The test statistic for the t-test is calculated using the formula: t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} Where:

  • xˉ\bar{x} = sample mean = 57
  • μ0\mu_0 = population mean under the null hypothesis = 55.1
  • ss = sample standard deviation = 3.44
  • nn = sample size = 23

Substitute the values: t=5755.13.44/23t = \frac{57 - 55.1}{3.44 / \sqrt{23}}

First, calculate the standard error (SE): SE=3.44233.444.7950.718SE = \frac{3.44}{\sqrt{23}} \approx \frac{3.44}{4.795} \approx 0.718

Now, calculate the t-value: t=5755.10.7181.90.7182.646t = \frac{57 - 55.1}{0.718} \approx \frac{1.9}{0.718} \approx 2.646

Therefore, the correct test statistic rounded to three decimal places is: t2.646t \approx 2.646

c) At α=0.1\alpha = 0.1, the rejection region:

  • We need to find the critical value for tt with α=0.1\alpha = 0.1 and degrees of freedom df=n1=231=22df = n - 1 = 23 - 1 = 22.

Using the t-distribution table or a calculator for t0.1,22t_{0.1, 22}, we find the critical value: t0.1,221.321t_{0.1, 22} \approx 1.321

So, the rejection region for this right-tailed test is: t>1.321t > 1.321

d) The decision is to:

  • Since our calculated tt-value 2.6462.646 falls in the rejection region t>1.321t > 1.321, we reject the null hypothesis H0H_0.

Therefore, the correct decision is: reject H0 since the test statistic falls in the rejection region.\text{reject } H_0 \text{ since the test statistic falls in the rejection region.}

Summary:

  • a) Correct
  • b) t=2.646t = 2.646
  • c) t>1.321t > 1.321
  • d) reject H0H_0

Would you like more details or have any questions about this hypothesis test?

Here are 8 related questions for further exploration:

  1. What is the formula for calculating the t-statistic in a one-sample t-test?
  2. How do you determine the degrees of freedom in a t-test?
  3. What is the significance level (α\alpha) in hypothesis testing, and how is it used?
  4. How do you interpret the p-value in the context of hypothesis testing?
  5. What are the assumptions underlying the use of a t-test?
  6. How does sample size affect the outcome of a t-test?
  7. What is the difference between a one-tailed and a two-tailed test?
  8. How do you use a t-distribution table to find critical values?

Tip: Always double-check your calculations and ensure you are using the correct formula and critical values for your hypothesis test.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
One-sample t-test
Significance Level
Critical Values

Formulas

t-statistic formula: t = (x̄ - μ₀) / (s / √n)

Theorems

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Suitable Grade Level

University