Answer:
A. The results contradict the belief that the mean body temperature is 98.6 °F because both the mean and the median are less than 98.6 °F.
Step-by-step explanation:
Assume the temperature data were those in the table below.
You would have calculated that
Mean = 97.9 °F
Median = 97.9 °F
These observations contradict the belief that the mean body temperature is 98.6 °F.
A 2017 study found the mean oral temperature of more than 35 000 British patients was 97.9 °F.
Answer:
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.
Step-by-step explanation:
Data given and notation
represent the sample mean
represent the sample standard deviation
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is less than 140, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
(1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
P-value
The first step is calculate the degrees of freedom, on this case:
Since is a one left tailed test the p value would be:
Conclusion
If we compare the p value and the significance level assumed we see that
so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.