Question

Medical examiners use temperatures of dead bodies to estimate time of death. This is especially important if a crime is suspected. Could the body be reheated to disguise the time of death? A biologist tested this issue in 1951 using mice. He studied the difference (reheated – freshly killed) in the cooling constants of each of the mice.

11) Why is this a paired differences study?

12) Write the hypotheses.

13) Fill in the third column and enter that into your calculator.

Freshly Killed	Reheated	Difference
573	481
482	343
377	383
390	380
535	454
414	425
438	393
410	435
418	422
368	346
445	443
383	342
391	378
410	402
433	400
405	360
340	373
328	373
400	412

[a] Using a 99% confidence interval for the mean difference in cooling constants, determine if there is significant evidence that reheating the bodies changes the cooling constant. (Remember that you just have one sample. Choose the Data option.)

[b] Calculate the P-value. Does a conclusion from the P-value strengthen or call into question your conclusion from the confidence interval? Explain. (Think about what you will use asthe “unusual sample” threshold.)

Answer 1

11) Why is this a paired differences study?

beacsuse sample remains same

12) Write the hypotheses.

Freshly Killed	Reheated	Difference
573	481	92
482	343	139
377	383	-6
390	380	10
535	454	81
414	425	-11
438	393	45
410	435	-25
418	422	-4
368	346	22
445	443	2
383	342	41
391	378	13
410	402	8
433	400	33
405	360	45
340	373	-33
328	373	-45
400	412	-12

a)

Paired T-Test and CI: Freshly Killed, Reheated

Paired T for Freshly Killed - Reheated

                 N   Mean StDev SE Mean
Freshly Killed 19 417.9   60.0     13.8
Reheated        19 397.1   39.1      9.0
Difference      19   20.8   45.8     10.5

99% CI for mean difference: (-9.5, 51.1)

since 0 is present in confidence interval , we fail to reject the null hypothesis

T-Test of mean difference = 0 (vs ? 0): T-Value = 1.98 P-Value = 0.064

b)

p-value = 0.064 > 0.01

hence we fail to reject the null