Prove that two confidence intervals overlap (t-distribution)

Question

I was able to solve part (a) but not part (b):

The temperature of a block of wood 3 minutes after being lifted out of liquid nitrogen is measured and then the experiment is repeated. The results are -1.2°C and 4.8°C.

a) Assuming that the temperatures are normally distributed, find a 95% confidence interval for the mean temperature of a block of wood 3 minutes after being lifted out of liquid nitrogen

$$X^- \pm t \biggl(\frac{s_{n-1}}{\sqrt{n}}\biggr) = 1.8 \pm 12.706 \biggl(\frac{\sqrt{18}}{\sqrt{2}}\biggr) = 1.8 \pm 12.706\cdot 3$$

where $t$ is the t-score associated with a 95% confidence interval when v=1.

Thus, the confidence interval is $[-36.3, 39,9]$.

b) A different block of wood is subjected to the same experiment and the results are 0°C and $x$°C where $x > 0$. Prove that the two confidence intervals overlap for all values of $x$.

I got that:

$$X^- \pm t \biggl(\frac{s_{n-1}}{\sqrt{n}}\biggr) = \frac{x}{2} \pm 12.706 \biggl(\frac{x/\sqrt{2}}{\sqrt{2}}\biggr) = \frac{x}{2}(1 ± 12.706).$$

But I don't think this proves that the two confidence intervals overlap...

Answer 1

Your calculations are correct. You must be misinterpreting the word "overlapping". It is not complete, but partial: $$-36.3<\frac{x}{2}(1-12.706) \ \ \text{OR} \ \ \frac{x}{2}(1+12.706)<39.9 \Rightarrow \\ x<6.15 \ \ \text{OR} \ \ x>5.82.$$ Thus, the intervals overlap for all $x>0$.

Answer 2

In part (a) it seem that $\bar X = 1.8$ and $S = 4.2426.$ Also, part of the output in R statistical software for a t test has the 95% CI

x = c(-1.2,4.8)
mean(x);  sd(x);  sqrt(18)
[1] 1.8
[1] 4.242641
[1] 4.242641

95 percent confidence interval:
 -36.31861  39.91861 

However, in part (b) it seems you have the wrong denominator for the sample variance and a wrong computation for the margin of error. Try data 0 and 12:

y = c(0, 12)
mean(y); sd(y)
[1] 6
[1] 8.485281