The manufacturer of the resistors claims that the expected maximum power at which the resistor fails is $0.25$.
You suspect this threshold is lower and therefore decide to measure the current and resistance at which the resistor breaks down.
By an ingenious application of Ohm’s Law this results in a sample mean of $\overline{x}_{20} = 0.21$ and a sample variance of $s^2_{20} = 0.01$ for the power.
You may assume that the measurements for the power are also normally distributed.
Use an appropriate statistical test to test the manufacturer’s claim at significance level $α=0.05$.

I assume that $H_0$: $\mu=0.25$ $H_1$: $\mu<0.25$



$P(Z\leq-1.788)=P(Z\geq1.788)= 0.0368$ (from the N(0,1) table)

I think that what I did is correct but I didn't use the value of $0.05$ for the significance level, how can I use it in this question? What I assumed is correct? Thanks for the help.

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    $\begingroup$ Compare the estimated probability ($0.0368$) with the criterion probability ($0.05$). $\endgroup$ – David G. Stork Jan 14 at 19:09
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    $\begingroup$ Or you compare the estimated probability with the criteria , as David mentioned above, or you could go to the table, obtain the Z-value for a probability of 0,05 and compare it with your Z . $\endgroup$ – RScrlli Jan 14 at 19:50
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    $\begingroup$ But this isn't a Z-test... $\endgroup$ – BruceET Jan 15 at 3:08

This is a t test (not a z test) because the population standard deviation $\sigma$ is unknown and estimated by the sample standard deviation $S = 0.1 = \sqrt{.01}.$

Your statement of null and alternative hypotheses is OK.

Here is output from Minitab statistical software for the t test:

One-Sample T 

Test of μ = 0.25 vs < 0.25

 N    Mean   StDev  SE Mean  95% Upper Bound      T      P
20  0.2100  0.1000   0.0224           0.2487  -1.79  0.045

What you call the $Z$-statistic is actually the $T$-statistic.

The P-value is the probability under Student's t distribution with degrees of freedom $\nu = n - 1 = 20-1 = 19$ to the left of $T = -1.79.$ It is difficult to find this P-value from printed tables of the t-distribution.

From a printed table you can find that the critical value for a left-sided test at the 5% level is -1.729. Because $T = -1.79 < -1.729,$ you know that you can reject at the 5% level.

From R statistical software, as below (also from Minitab and others), one can find the P-value which agrees with the value in the printout: 0.045.

qt(.05, 19)
[1] -1.729133
pt(-1.79, 19)
[1] 0.04470223

Note: Student's t distribution with $\nu = 19$ is "close" to a standard normal distribution, but not exactly. If this were a z-test, then the critical value would be -1.645 and the P-value would be 0.367, as you say.

[1] -1.644854
[1] 0.03672696

It may be inevitable that every beginning statistics student must confuse a t-test with a z-test once during the first course. Perhaps this can be your one such mistake, avoiding another on your next exam.

Below are plots of the density of $\mathsf{T}(19)$ (solid blue) and the density of $\mathsf{Norm}(0,1)$ (dotted). The vertical red line is at $T = -1.79.$ The P-value of the t test is the area under the blue curve to the left of this red line. The correct 5% critical value for the t distribution is shown by the solid blue line.

enter image description here

  • $\begingroup$ Thank you a lot for the answer and for your time BruceET, you are right I didn't recognize that it was a t-test, but now it's clear the reason, thanks again! $\endgroup$ – FTAC Jan 15 at 7:47

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