recent suggestion is 60 min of exercise per day. A sample of 50 adults in a study of risk factor report exercising a mean of 38 min per day with a standerd deviation of 19 min .based on sampal data, is physical activity significantly less than recommended? Run the appropriate test at a 5% level of significance.
Use a one-sample z-test.
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (38 - 60)/(19/√50) = ?
Finish the calculation.
Check a z-table at .05 level of significance for a one-tailed test.
If the z-test statistic exceeds the critical value from the z-table, reject the null and conclude µ < 60 (which is the alternative hypothesis). If the z-test statistic does not exceed the critical value from the z-table, do not reject the null.
I hope this will help get you started.
To determine if physical activity is significantly less than the recommended 60 minutes per day, we can use a one-sample t-test. Here's how you can calculate and interpret the results:
Step 1: State the hypotheses
In this case, we want to determine if the mean exercise time is significantly less than 60 minutes per day.
Null hypothesis (H0): The mean exercise time is equal to 60 minutes per day.
Alternative hypothesis (HA): The mean exercise time is less than 60 minutes per day.
Step 2: Set the significance level
The significance level (alpha) represents the threshold for concluding whether there is sufficient evidence to reject the null hypothesis. In this case, the significance level is 5% or 0.05.
Step 3: Collect and summarize the data
From the sample of 50 adults, we have:
Sample mean (x̄) = 38 minutes per day
Sample standard deviation (s) = 19 minutes per day
Sample size (n) = 50
Step 4: Calculate the test statistic
The test statistic for a one-sample t-test is calculated using the formula:
t = (x̄ - µ) / (s / sqrt(n))
Where:
x̄ = sample mean
µ = hypothesized population mean (60 minutes in this case)
s = sample standard deviation
sqrt = square root
n = sample size
Plugging in the values from the given data, we get:
t = (38 - 60) / (19 / sqrt(50))
Step 5: Determine the critical value and p-value
Using a t-distribution table or a statistical software, find the critical value for a one-tailed test at a 5% significance level with n-1 degrees of freedom (49 in this case).
Alternatively, you can find the p-value using the t-distribution.
Step 6: Make a decision
If the test statistic is greater than the critical value (or if the p-value is less than the significance level α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, if the calculated test statistic falls in the rejection region (either above the critical value or with a p-value less than 0.05), you would conclude that the physical activity is significantly less than the recommended 60 minutes per day.
Note: Since this response cannot calculate the test statistic, you will need to perform the calculations either manually or using statistical software to obtain the actual test statistic and make the decision.