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 a 0.05 for both, is there sufficient evidence that
these two results differ from the national means?
Instructions: Fill in the required information on the blanks provided after each question. Use the t-table provided in the module to solve for critical values. If the test is two-tailed, write '+-' before the values to denote i,e. if the critical values are t=-1.098 and t=1.098, simply write +-1.098. For numerical answer, round up to three decimal places
Men spend an average of 30 minutes per day on weekends and holidays exercising and playing sports. A random sample of 20 men resulted in a mean of 34 minutes exercising with a standard deviation of 6 minutes. At a = 0.05, is there sufficient evidence that these results differ from the national mean?
1. The critical value for the t-test is t=
2. The value of the t-test statistic obtained from the sample is t=
3. (Write R to Reject Ho, or N to NOT Reject Ho) The Decision therefore is to
4. (Write Y for Yes, or N for No) Is there enough evidence to support the claim?
ddend
To determine if there is sufficient evidence that the sample means differ from the national means, we can perform a two-sample t-test.
Let's denote the following variables:
- x̄1 = sample mean for exercise time (35 minutes)
- s1 = sample standard deviation for exercise time (6.9 minutes)
- n1 = sample size for exercise time (25)
- x̄2 = sample mean for reading time (20.5 minutes)
- s2 = sample standard deviation for reading time (7.2 minutes)
- n2 = sample size for reading time (25)
- μ1 = population mean for exercise time (29 minutes)
- μ2 = population mean for reading time (23 minutes)
- α = significance level (0.05)
The null hypothesis (H0) is that there is no difference between the sample means and the population means. The alternative hypothesis (H1) is that there is a difference between the sample means and the population means.
We can calculate the test statistic using the formula:
t = ((x̄1 - x̄2) - (μ1 - μ2)) / sqrt((s1^2 / n1) + (s2^2 / n2))
Now let's plug in the values and calculate the t-statistic:
t = ((35 - 20.5) - (29 - 23)) / sqrt((6.9^2 / 25) + (7.2^2 / 25))
Next, we need to determine the degrees of freedom (df) for the t-distribution. Since both sample sizes are equal (n1 = n2 = 25), we can use the pooled variance formula:
df = n1 + n2 - 2 = 25 + 25 - 2 = 48
Using a t-table or a t-distribution calculator, we can determine the critical t-value for a two-tailed test with a significance level of 0.05 and 48 degrees of freedom.
Let's assume the critical t-value is 2.010. If the calculated t-statistic is greater than the critical t-value or less than the negative of the critical t-value, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Calculate the t-statistic using the given values, and compare it to the critical t-value to determine whether or not there is sufficient evidence to reject the null hypothesis.
To answer this question, we will perform a two-sample t-test to determine if there is sufficient evidence that the mean exercise time and mean reading time of the sample differ significantly from the national means.
Here are the steps to perform the t-test:
Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) assumes that there is no significant difference between the sample means and the national means.
The alternative hypothesis (Ha) assumes that there is a significant difference between the sample means and the national means.
H0: μ_exercise = 29 (national mean exercise time)
Ha: μ_exercise ≠ 29
H0: μ_reading = 23 (national mean reading time)
Ha: μ_reading ≠ 23
Step 2: Determine the significance level (α):
The significance level (α) given in the question is 0.05. This represents the probability of rejecting the null hypothesis when it is true.
Step 3: Compute the test statistic:
We will use the formula for the two-sample t-test:
t = (x1 - x2 - (μ1 - μ2)) /√((s1^2 / n1) + (s2^2 / n2))
Where:
x1 = sample mean of exercise time
x2 = sample mean of reading time
μ1 = national mean of exercise time
μ2 = national mean of reading time
s1 = standard deviation of exercise time
s2 = standard deviation of reading time
n1 = sample size of exercise time
n2 = sample size of reading time
Substituting the given values:
x1 = 35
x2 = 20.5
μ1 = 29
μ2 = 23
s1 = 6.9
s2 = 7.2
n1 = n2 = 25 (since we have a sample of 25 men for each variable)
t = (35 - 20.5 - (29 - 23)) /√((6.9^2 / 25) + (7.2^2 / 25))
Step 4: Determine the critical value(s):
Since we are conducting a two-tailed test, we divide the significance level (α) by 2, resulting in α/2 = 0.025. We will use this value to find the critical t-value from the t-distribution table for the given degrees of freedom (df = n1 + n2 - 2).
Step 5: Compare the test statistic with the critical value(s):
If the absolute value of the calculated test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is sufficient evidence to support the alternative hypothesis. If it is less than the critical value, we fail to reject the null hypothesis.
Step 6: Calculate the p-value:
If the test statistic falls outside the critical region, we can calculate the p-value associated with it. The p-value represents the probability of obtaining the observed sample difference (or more extreme) under the null hypothesis.
Step 7: Make a conclusion:
Based on the comparison of the test statistic with the critical value(s) or the p-value, we can make a conclusion and determine if there is sufficient evidence to support the alternative hypothesis.
Performing these calculations will allow us to determine if the results for exercise and reading times in the sample differ significantly from the national means.