Follwing data desribes the measured fracture strength of 15 randomly selected units made by a new ceramic material $$20, 42, 18, 21, 22, 35, 19, 18, 16, 20, 21, 32, 22, 20, 24$$ At a previously materials used, the fracture strength haad meadian value 25. Test with $5 \%$ significane level if the new materials fracture strength have lower meadian.
My attempt(need comments about if i have done right or not etc)
My first thought was to decide if I could use the method for sample in pairs or not. Since we cannot suppose that this is normally distributed and since we have $np^2_0<5$ we cannot use normalapproximation either. therefore we need to use the method of sign test. From the bookexercise above we can see that we want to test the null hypothesis $$H_0: \bar m = 25$$ against the alternative hypothesis $$H_1: \bar m < 25$$ because we want to see if the new material have lower median. We let $z_i=x_i - \bar m_0$ be the differences of observations and the median.If $u_{-}=$ number of f $z_i<0$ then $u_{-}$ ($u_-$ is the testvariable?) is an observation of $U_- \sim Bin(15, 1/2)$. By the direct method and that $u_- = 3$ we get that if the $P-value$ is lower than the significance level $5 \%$ then $H_0$ is rejected:
$P=P_{H_0}$(have atleast one equally extreme outcomes as the observed)$=P_{H_0}(U_- \leq 3)=0.0176$.
We get that $$P<0.05$$ and therefore the null hypothesis is rejected. This means that the new material has a lower median.
Number of negative differences are $12$ and positive are $3$. Therefore the equally extreme outcomes as the observed should be $U_- \leq 3$ given that $H_0$ is true. Equally we could use that $U_- \geq 12$ and get same answer ( my assumption. If this is true, could someone tell me why it's like that?)
Extra question: The rejection region should be of the type $\{u \leq K \}$ right? Should $K=3$?
This looks mainly correct (some of the formatting is garbled as I read it on my computer.) I will go through the logic of the sign test in some detail because you indicate you don't fully understand what is going on.
You want to test $H_0: \eta = 25$ against alternative $H_1: \eta < 25,$, where we use small Greek $\eta$ (eta) for the unknown population median fracture strength.
If the null hypothesis is true, half of your observations will tend to lie below $\eta = 25$ and half above. So we count the number of your 15 observations below. I count 12 below and 3 above, which agrees with your count. (If any observation happened to be exactly equal to 25, it would be removed from the data for purposes of this test.)
We designate by $U$ the number of observations below. Under the null hypothesis $U \sim Binom(n = 15, p = .5).$ One could consider this the null distribution of a test of the null hypothesis that $p = 1/2,$ against the alternative that $ p > 1/2$.
Equivalently, would you consider a coin biased against Heads if you got only 3 Heads in 15 tosses? (This is also the same as asking if a coin is biased for Tails if you get 12 Tails in 15 tosses.)
The P-value is $P\{U \le 3 | p=.5\} = 0.01757813 < 0.05,$ as you say.
If you are doing this in terms of a fixed significance level $\alpha = .05$, and you seek the critical value $K$ for a rejection region of the form $\{U \le K\},$ then $K = 3$ because $P\{U \le 4\} = 0.0592$, which is above 5%, and $P\{U \le 3\} = 0.0175$, which is below 5%.
Note: In general, when the distribution of the test statistic is discrete, it is not possible to have a test at EXACTLY the 5% level. Here you have choice between $\alpha = 5.92\%$ and $\alpha = 1.75\%$, and choose the latter.
Addendum: A dataset for which the sign test is superior to Wilcoxon signed rank or t-test: You say this is a textbook problem on the sign test. It is an excellent example of a dataset for which he sign test is appropriate. Usual alternatives might be the Wilcoxon signed rank test, inappropriate here because there are many ties (only 10 uniquely different values among 15); of the one-sample t-test, inappropriate here because of severe right-skewing (the three values above 25 might all be considered outliers and a Shapiro-Wilk test overwhelmingly rejects normality with P-value 0.0019). When applied in spite of contraindications, neither of these tests rejects the null hypothesis at the 5% level. I am 'upvoting' this question because of its potential usefulness as a pedagogical example on selection of tests.