Test the claim that the mean body temperature is less than 98.6 using a= 0.05. Assume a smaple of 16 yilded a mean of 98.2 with a standard deviation of 0.6
Z = difference between means/√SE^2
SE = SD/√(n-1) (or just √n)
Look up value in table in back of statistics text labeled something like "areas under normal distribution.
I hope this helps. Thanks for asking.
To test the claim that the mean body temperature is less than 98.6, we can use a one-sample t-test. The steps involved in conducting the t-test are as follows:
Step 1: State the hypotheses:
- Null hypothesis (H0): The mean body temperature is equal to or greater than 98.6.
- Alternative hypothesis (Ha): The mean body temperature is less than 98.6.
Step 2: Set the significance level (α):
Since a significance level of α = 0.05 is given in the question, we will use that value.
Step 3: Compute the test statistic:
The test statistic for a one-sample t-test is calculated using the following formula:
t = (x̄ - μ) / (s / √n)
Where:
- x̄ is the sample mean (98.2)
- μ is the population mean under the null hypothesis (98.6)
- s is the sample standard deviation (0.6)
- n is the sample size (16)
Plugging in the values, we get:
t = (98.2 - 98.6) / (0.6 / √16)
t = -0.4 / (0.6 / 4)
t ≈ -0.4 / 0.15
t ≈ -2.67
Step 4: Determine the critical value(s):
Since the alternative hypothesis is that the mean body temperature is less than 98.6, we are conducting a one-tailed test and the critical value can be found in the t-distribution table. At a significance level of 0.05 and degrees of freedom (df = n-1 = 15), the critical value is approximately -1.753.
Step 5: Make a decision:
If the test statistic falls in the critical region (i.e., if t < -1.753), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In our case, since t = -2.67 falls in the critical region (t < -1.753), we reject the null hypothesis.
Step 6: Interpret the result:
Based on the sample data, we have enough evidence to conclude that the mean body temperature is less than 98.6 at a significance level of 0.05.