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If you throw a coin in a vending machine, the coin is being weighed by the machine to determine its value.
For statistical purposes, you decide to throw $4$ fifty-cent coins in this machine and let $\overline{X}_4$ be the estimator for the mean weight $μ$ of a fifty-cent coin.
Assume that a single measurement has a normal distribution with expectation $μ$ and variance $σ^2 = 0.04$.
You find out that $\overline{x}_4 = 7.54$.
Investigate by using a statistical test whether $μ$ equals $7.5$ at significance level $α = 0.01$.

Solve:
I assume $H_0:\mu=7.5$ and $H_1:\mu\neq7.5$ So it is a two-tailed problem and I have an $\alpha=0.005$ on the right and the same on the left. So from the table the critical values for these $\alpha$ are 2.57 and -2.57. So I can compute $Z=\frac{7.48-7.5}{\frac{\sqrt{0.004}}{\sqrt{4}}}=-0.63$ that is inside the region where we don't reject $H_0$ so we don't reject it.
Is it correct? Can someone please help me? Or I should use the p-value?

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You have written the population variance as $\sigma^2 = 0.004,$ not $0.04.$ Also, in the problem you state that $\bar X = 7.54,$ but in your computation you have $7.48.$ Otherwise, you seem to be on the right track.

After fixing your typos, you should have $Z = 0.4.$ Then at level $\alpha = 0.01 = 1\%,$ you would reject if $|Z| > 2.576,$ so you cannot reject.

The P-value is the area under the standard normal curve outside the interval $(-.4,.4),$ which is $0.6892 > 0.01,$ and so (again) no rejection.

My computations from R statistical software are shown below:

(7.54-7.50)/ sqrt(.04/4)
[1] 0.4
qnorm(.995)
[1] 2.575829
2*pnorm(-.4)
[1] 0.6891565

Also, Minitab statistical software gives the following relevant output:

One-Sample Z 

Test of μ = 7.5 vs ≠ 7.5
The assumed standard deviation = 0.2

N   Mean  SE Mean      99% CI         Z      P 
4  7.540    0.100  (7.282, 7.798)  0.40  0.689

Because your hypothetical mean $\mu = 7.5$ is included in the 99% confidence interval, you would not reject at the 1% level.

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    $\begingroup$ Ohh yes, I see my errors, thank you, so I'll reject the null hypothesis when $Z>0.4$ and when $Z<-0.4$, because both are critical values, correct? And the p-value should be $P(Z>0.4)+P(Z<-0.4)$? $\endgroup$ – Mark Jacon Jan 27 at 11:38
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    $\begingroup$ Both comments correct. Also reject at 1% if 99% CI does not include hypothetical mean. 99% CI when $\sigma$ is known is of form $\bar X \pm 2.576\sigma/\sqrt{n}.$ Try it and compare with CI in Minitab output. $\endgroup$ – BruceET Jan 27 at 19:29

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