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
Did I solve this correctly?
a. H0: 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
Test statistic = (-0.35-0)/0.3/โ4=-2.333
c. 0.051
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 6pm is higher than the body temperature measured at 6am.
Your analysis seems correct. I will go through each step to verify:
a. The symbolic null and alternative hypothesis are:
H0: ยตd โฅ 0 (The body temperature measured at 6 pm is equal to or greater than the body temperature measured at 6 am)
HA: ยตd < 0 (The body temperature measured at 6 pm is lower than the body temperature measured at 6 am)
b. To determine the test statistic, you first need to calculate the differences between the body temperatures measured at 6 am and 6 pm:
D = (98.0 - 98.0), (97.0 - 97.6), (98.6 - 98.8), (97.4 - 98.0)
= 0, -0.6, -0.2, -0.6
Next, calculate the sample size, average, and standard deviation of the differences:
Sample size (n) = 4
Average (๐ฅฬ) = (0 + -0.6 + -0.2 + -0.6) / 4 = -0.35
Standard Deviation (s) = โ((0 - (-0.35))^2 + (-0.6 - (-0.35))^2 + (-0.2 - (-0.35))^2 + (-0.6 - (-0.35))^2) / (4-1) = 0.3
Now, calculate the test statistic, which is the average difference (๐ฅฬ) divided by the standard deviation of the differences (s) divided by the square root of the sample size (n):
Test statistic = (-0.35 - 0) / (0.3 / โ4) = -2.333 (rounded to 3 decimal places)
c. To determine the p-value, you need to find the probability of observing a t-value less than the test statistic (-2.333) with degrees of freedom (df) equal to n-1. In this case, df = 4 - 1 = 3.
Using a t-distribution table or a t-distribution calculator, you can find that the p-value is approximately 0.051 (rounded to 3 decimal places).
d. Since the p-value (0.051) is smaller than the significance level of 0.10, we reject the null hypothesis (H0).
e. Therefore, based on the data, we have enough evidence to conclude that the body temperature measured at 6 pm is lower than the body temperature measured at 6 am.
Your answer is almost correct, but there is a small mistake in part d:
d. Result:
Reject H0
Reason: The p-value of 0.051 is greater than the significance level of 0.10.
Since the p-value is greater than the significance level, we fail to reject the null hypothesis.
Therefore, the correct conclusion is:
We fail to reject the null hypothesis. There is not enough evidence to support the claim that the body temperature measured at 6 pm is higher than the body temperature measured at 6 am.