Math Problem Statement

Students estimated the length of one minute without reference to a watch or​ clock, and the times​ (seconds) are listed below. Assume that a simple random sample has been selected. Use a 0.10

significance level to test the claim that these times are from a population with a mean equal to 60

seconds. Does it appear that students are reasonably good at estimating one​ minute? 70 83 39 66 41 21 60 62

65 47 66 69 95 91 67

Question content area bottom Part 1 Perform the test assuming that the requirements are met. Identify the null and alternative hypotheses. Upper H 0 ​: mu equals 60Upper H 1 ​: mu not equals 60 ​(Type integers or decimals. Do not​ round.) Part 2 Identify the test statistic. enter your response here ​(Round to two decimal places as​ needed.)

Solution

To test the claim that the times are from a population with a mean equal to 60 seconds, we'll conduct a one-sample t-test. Here's the step-by-step solution:

Part 1: Hypotheses

The null and alternative hypotheses are:

  • Null Hypothesis (H0H_0): μ=60\mu = 60 (The mean time is equal to 60 seconds.)
  • Alternative Hypothesis (H1H_1): μ60\mu \neq 60 (The mean time is not equal to 60 seconds.)

Part 2: Test Statistic

First, we need to calculate the sample mean (xˉ\bar{x}) and the sample standard deviation (ss).

Given the dataTo test the claim that the times are from a population with a mean equal to 60 seconds, we'll conduct a one-sample t-test. Here's the step-by-step solution:

Part 1: Hypotheses

The null and alternative hypotheses are:

  • Null Hypothesis (H0H_0): μ=60\mu = 60 (The mean time is equal to 60 seconds.)
  • Alternative Hypothesis (H1H_1): μ60\mu \neq 60 (The mean time is not equal to 60 seconds.)

Part 2: Test Statistic

First, we need to calculate the sample mean (xˉ\bar{x}) and the sample standard deviation (ss).

Given the 70,83,39,66,41,21,60,62,65,47,66,69,95,91,6770, 83, 39, 66, 41, 21, 60, 62, 65, 47, 66, 69, 95, 91, 67

Step 1: Calculate the sample mean (xˉ\bar{x})

xˉ=xin\bar{x} = \frac{\sum x_i}{n} Where:

  • n=15n = 15 (the number of observations)
  • xi\sum x_i is the sum of all observations.

Step 2: Calculate the sample standard deviation (ss)

s=(xixˉ)2n1s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

Step 3: Calculate the test statistic (t)

t=xˉμ0s/nt = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} Where:

  • μ0=60\mu_0 = 60 (the population mean under the null hypothesis)

I'll now calculate these values.### Results:

  • Sample Mean (xˉ\bar{x}): 62.8 seconds
  • Sample Standard Deviation (ss): 19.68 seconds (rounded to two decimal places)
  • Test Statistic (t): 0.55 (rounded to two decimal places)

Conclusion:

Using the calculated test statistic of t=0.55t = 0.55, you can now compare this with the critical value from the t-distribution table for a two-tailed test at the 0.10 significance level with n1=14n-1 = 14 degrees of freedom, or directly find the p-value associated with this t-statistic.

Do you want to proceed with finding the p-value or critical value to finalize the hypothesis test? Let me know if you need further details!


Here are five related questions you might want to explore:

  1. How do you determine the degrees of freedom in a t-test?
  2. What is the p-value, and how do you interpret it?
  3. How does the significance level affect the outcome of a hypothesis test?
  4. What assumptions must be met for the t-test to be valid?
  5. How would the test results change if the sample size were larger?

Tip: When conducting a hypothesis test, always ensure the assumptions (e.g., normality, independence) are reasonably met to validate your results.

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Math Problem Analysis

Mathematical Concepts

Statistics
Hypothesis Testing
T-Test

Formulas

Sample Mean: \(\bar{x} = \frac{\sum x_i}{n}\)
Sample Standard Deviation: \(s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\)
T-Test Statistic: \(t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\)

Theorems

Central Limit Theorem
T-Distribution

Suitable Grade Level

Undergraduate Statistics