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An agricultural researcher plants in a random sample of $25$ plots with a new variety of corn. The average yield for these plots is $\bar{x} = 150$ bushels per acre with a standard deviation $s = 10$ bushels per acre . Assume that the yield per acre for the new variety of corn follows a Normal distribution with unknown mean µ and standard deviation σ. Calculate A $95\%$ confidence interval for $u$

my attempt

$n = 25, \bar{x} = 150, s = 10, a = 0.95$. Keyword: Unknown mean $u$ means we use $t$ table.

$$\bar{x} \pm t_{\frac{1+0.95}{2}} \frac{s}{\sqrt{n}}$$

$$150 \pm t_{0.975} \frac{10}{\sqrt{25}}$$

$$150 \pm t_{0.975} \times 2$$

But this is wrong according to the solution:

$$150 \pm t_{0.95} \times 2$$

How?

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  • $\begingroup$ The fact that $\sigma$ is unknown indicates that you have to use '$t$ table' (mean is of course unknown, hence you are estimating the mean). And the 't' fractile is supposed to be $0.05/2=0.025$. $\endgroup$ – StubbornAtom Apr 12 '19 at 19:00
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I agree with you. The confidence interval is

$$\large{\left[\overline x-t_{(1-\tfrac{\alpha}{2};n-1)}\cdot \frac{s}{\sqrt n} ; \ \overline x+t_{(1-\tfrac{\alpha}{2};n-1)}\cdot \frac{s}{\sqrt n} \right]}$$

In your case $1-\tfrac{\alpha}{2}=1-\frac{0.05}{2}=1-0.025=0.975, n=25, \overline x=150$ and $s=10$. Thus the interval is

$\left[150-t_{(0.975;24)}\cdot \frac{10}{5} ; \ 150+t_{(0.975;24)}\cdot \frac{10}{5} \right]$, where $t_{(0.975;24)}=2.064$

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  • $\begingroup$ The lower and upper limits of the confidence interval should be swapped. $\endgroup$ – StubbornAtom Apr 12 '19 at 19:15
  • $\begingroup$ @StubbornAtom I don´t think so. The lower limit is smaller than the upper limit. At my interval I have $t_{(1-\tfrac{\alpha}{2};n-1)}$ and not $t_{(\tfrac{\alpha}{2};n-1)}$ $\endgroup$ – callculus Apr 12 '19 at 19:18
  • $\begingroup$ Because $t_{0.025;24}=t_{1-0.975;24}=-t_{0.975;24}=2.064$. $\endgroup$ – StubbornAtom Apr 12 '19 at 19:21
  • $\begingroup$ @StubbornAtom No, it is $t_{0.975,24}=+2.064$. $\endgroup$ – callculus Apr 12 '19 at 19:24
  • $\begingroup$ @StubbornAtom Nice, that we have a common sense. $\endgroup$ – callculus Apr 12 '19 at 19:33

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