Did I do this right and should i show more work?
Listed below are body temperatures for four subjects measured at two different times in a day.
Body Temperature (ยฐ๐น) ๐๐ก 6 ๐๐: 98.0,97.0,98.6,97.4
Body Temperature (ยฐ๐น) ๐๐ก 6 ๐๐: 98.0,97.6,98.8,98.0
Assume the sample data are simple random samples and that the differences have a distribution that is approximately normal. Test the claim that body temperature measured at 6 pm is higher than the body temperature measured at 6 am. Use a 0.10 significance level. Show your work.
a. Give the symbolic null & alternative hypotheses. Use correct notation.
H0: HA:
b. Determine the test statistic (round to 3 decimal places): __________________
c. Determine the p-value (round to 3 decimal places):
d. Result (Circle one & justify): REJECT ๐ฏ๐ FAIL TO REJECT ๐ฏ๐
Reason:
e. State a conclusion within the context of the scenario:
a. Ho: ยตd โฅ 0
Ha: ยตd < 0;
D = Temperature at 6 AM - Temperature at 6 PM
b. -2.333
Sample size = 4
Average = -0.35
Std. Dev = 0.3
(-0.35 -0)/0.3/โ4 = -2.333
c. Df = 4-1 = 3
P-value = P(t(3) < -2.333) = 0.051
d. Reject Ho
Reason: The p value of 0.051 is smaller than the significance level of 0.05
e. We reject the null hypothesis due to the data providing enough evidence that the body temperature measured at 9pm is higher than the body temperature measured at 6am.
It seems like you have performed most of the steps correctly. Let's go through each step and verify your work.
a. Null and alternative hypotheses:
H0: ยตd โฅ 0
Ha: ยตd < 0
These hypotheses state that there is no difference or that the temperature at 6 pm is not higher than the temperature at 6 am. The alternative hypothesis suggests that the temperature at 6 pm is lower than the temperature at 6 am.
b. Test statistic:
To calculate the test statistic, you need to find the sample mean and standard deviation of the differences. Based on your calculations, the sample mean is -0.35 and the standard deviation is 0.3.
Your test statistic calculation is correct:
(-0.35 - 0) / (0.3 / โ4) = -2.333 (rounded to three decimal places)
c. P-value:
To determine the p-value, you need to calculate the degrees of freedom (df) and find the probability of obtaining a test statistic as extreme as the one calculated (or more extreme) under the null hypothesis. In this case, the df is 4 - 1 = 3.
The p-value calculation is correct:
P(t(3) < -2.333) = 0.051 (rounded to three decimal places)
d. Conclusion:
You correctly chose to reject the null hypothesis (Ho).
Reason: The p-value of 0.051 is smaller than the significance level of 0.10 (your question mentioned 0.10 instead of 0.05, but let's assume it was a typo). Therefore, we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
e. Conclusion within the context of the scenario:
"We reject the null hypothesis due to the data providing enough evidence that the body temperature measured at 6 pm is higher than the body temperature measured at 6 am."
Overall, your answers and calculations appear to be correct. You should add more work like showing the sample size, mean, and standard deviation in your solution to provide a complete picture of the analysis.