Men spend an average of 29 minutes per day on weekends and holidays exercising and playing sports. They spend an average of 23 minutes per day reading. A random sample of 25 men resulted in a mean of 35 minutes exercising with a standard deviation of 6.9 minutes and an average of 20.5 minutes reading with s = 7.2 minutes. At 0.05 for both, is there sufficient evidence that these two results differ from the national means? Find the P-value.
To determine if there is sufficient evidence that these two results differ from the national means, we can perform two independent t-tests. Let's start by calculating the t-statistic and p-value for the exercise data.
1. For exercise data:
Population mean (μ): 29 minutes
Sample mean (x̄): 35 minutes
Sample size (n): 25
Sample standard deviation (s): 6.9 minutes
The null hypothesis (H0) is that the sample mean is equal to the population mean:
H0: x̄ = μ
The alternative hypothesis (Ha) is that the sample mean is not equal to the population mean:
Ha: x̄ ≠ μ
To calculate the t-statistic, we use the formula:
t = (x̄ - μ) / (s / √n)
Substituting the values:
t = (35 - 29) / (6.9 / √25)
t = 6 / (6.9 / 5)
t = 6 / 1.38
t ≈ 4.34
2. For reading data:
Population mean (μ): 23 minutes
Sample mean (x̄): 20.5 minutes
Sample size (n): 25
Sample standard deviation (s): 7.2 minutes
The null hypothesis (H0) is: x̄ = μ
The alternative hypothesis (Ha) is: x̄ ≠ μ
Using the same formula, we calculate the t-statistic for reading data:
t = (20.5 - 23) / (7.2 / √25)
t = -2.5 / (7.2 / 5)
t = -2.5 / 1.44
t ≈ -1.74
To determine whether there is sufficient evidence, we need to compare the calculated t-values with the critical t-value at the specified significance level (α = 0.05).
Looking up the critical t-value for a two-tailed test with 24 degrees of freedom (n-1), we find:
- Critical t-value = ±2.064
For the exercise data:
Since |t| = 4.34 > 2.064, we reject the null hypothesis.
For the reading data:
Since |t| = 1.74 < 2.064, we fail to reject the null hypothesis.
Now, let's calculate the p-values for both tests.
For the exercise data, the p-value is the probability that the t-statistic is as extreme as the observed value (or even more extreme) under the null hypothesis.
Using a t-distribution table or a calculator, we find that the p-value is very small, approximately < 0.001.
For the reading data, the p-value is also the probability that the t-statistic is as extreme as the observed value (or even more extreme) under the null hypothesis.
Using a t-distribution table or a calculator, we find that the p-value is approximately 0.093.
In conclusion:
- For the exercise data, there is sufficient evidence to suggest that the sample mean differs from the national mean.
- For the reading data, there is insufficient evidence to suggest that the sample mean differs from the national mean.
To determine if there is sufficient evidence that the results differ from the national means, we can perform a hypothesis test for the means of two independent samples.
Hypotheses:
Null Hypothesis (H₀): There is no significant difference between the sample means and the national means.
Alternative Hypothesis (H₁): There is a significant difference between the sample means and the national means.
We will conduct two separate hypothesis tests for the mean exercise time and the mean reading time.
1. Mean Exercise Time:
Given:
Sample mean (x̄1) = 35 minutes
Sample standard deviation (s1) = 6.9 minutes
Sample size (n1) = 25
Population mean (μ1) = 29 minutes (from national average)
To determine the test statistic, we need to calculate the t-score:
t = (x̄1 - μ1) / (s1 / sqrt(n1))
t = (35 - 29) / (6.9 / sqrt(25))
t = 6 / 1.38
t ≈ 4.35
Using a t-distribution table or a calculator, we find the critical value for a two-tailed test at α = 0.05 and degrees of freedom (df) = n1 - 1 = 25 - 1 = 24.
The critical t-value is approximately ±2.064.
Since the calculated t-value (4.35) is greater than the critical t-value (±2.064), we can reject the null hypothesis for the mean exercise time. There is sufficient evidence to suggest that the mean exercise time differs from the national mean.
To find the p-value for this test, we can calculate the area under the t-distribution curve that is more extreme than our observed t-value.
Using a t-distribution table or a calculator, we find that the p-value for a t-value of 4.35 with 24 degrees of freedom is less than 0.001 (almost 0.0001).
2. Mean Reading Time:
Given:
Sample mean (x̄2) = 20.5 minutes
Sample standard deviation (s2) = 7.2 minutes
Sample size (n2) = 25
Population mean (μ2) = 23 minutes (from national average)
To determine the test statistic, we need to calculate the t-score:
t = (x̄2 - μ2) / (s2 / sqrt(n2))
t = (20.5 - 23) / (7.2 / sqrt(25))
t = -2.5 / 1.44
t ≈ -1.74
Using the t-distribution table or a calculator, we find the critical value for a two-tailed test at α = 0.05 and degrees of freedom (df) = n2 - 1 = 25 - 1 = 24.
The critical t-value is approximately ±2.064.
Since the calculated t-value (-1.74) is less than the critical t-value (±2.064), we fail to reject the null hypothesis for the mean reading time. There is insufficient evidence to suggest that the mean reading time differs significantly from the national mean.
To find the p-value for this test, we can calculate the area under the t-distribution curve that is more extreme than our observed t-value.
Using a t-distribution table or a calculator, we find that the p-value for a t-value of -1.74 with 24 degrees of freedom is approximately 0.0955.
In conclusion, there is sufficient evidence to suggest that the mean exercise time is different from the national mean, as the p-value for the exercise test is less than the significance level of 0.05. However, there is insufficient evidence to suggest that the mean reading time significantly differs from the national mean, as the p-value for the reading test is greater than the significance level of 0.05.