A parameter is a scalar or vector indexing a family of probability distributions.
A pivotal quantity is a function of a statistic and the parameter of interest, and follows a standard distribution. The distribution may not involve any unknown parameters.
In parametric statistical inference, an hypothesis is a statement about the parameters of a probability distribution.
A simple hypothesis specifies a single value for a parameter.
P-value is the smallest level of significance that leads to rejection of the null hypothesis, or the largest level of significance that leads to rejection, for the given data. It is also the probability of getting a value of the test statistic as extreme as, or more extreme than, the value observed, if the null hypothesis were true, where the direction of "extreme" is determined by the alternative hypothesis.
n = 6, xbar = 17.3, s2 = 2.8
The hypothesis is that the mean is 15 MPa, the alternative is 2-sided, and we assume that s = 4.
> z0 <- (17.3 - 15)/(4/sqrt(6)) > z0 [1] 1.408457 > qnorm(.975) [1] 1.959964
The Reference Distribution is N(0,1). The 2-sided 5% critical values are +/-1.960, so we do not reject the hypothesis that the mean is 15 MPa at the 5% level of significance.
Assume that the observations are independent and come from a distribution that is at least approximately normal.
A 95% confidence interval for the standard deviation is
> sqrt(c(5*2.8/qchisq(.975,5), 5*2.8/qchisq(.025,5))) [1] 1.044500 4.104008
Since the hypothesised value s = 4 lies in this interval, we do not reject the null hypothesis at the 5% level of significance.
Assume that the observations are independent and come from a distribution that is exactly normal.
The hypothesis is that the mean is 15 MPa, the alternative is 2-sided, and we estimate s by s.
> t0 <- (17.3 - 15)/sqrt(2.8/6) > t0 [1] 3.366855 > qt(.975,5) [1] 2.570582
The Reference Distribution is t(5). The 2-sided 5% critical values are +/-2.571, so we reject the hypothesis that the mean is 15 MPa at the 5% level of significance.
Assume that the observations are independent and come from a distribution that is exactly normal.
> ((qnorm(.995) + qnorm(.9))*(0.012)/0.01)^2 [1] 21.42632
This is a straightforward application of the sample size formula for a two-sided test for the mean of a normal distribution. Rounding up gives that n = 22 observations are required.
Let X be the number of heads in 6 tosses. Assuming that the tosses are independent, X ~ Bin(6, p). The hypothesis is that p = 0.5, the alternative is two-sided. Since we observed x = 6, which is in the right tail of the reference distribution, the P-value is twice the right tail probability.
> 2*(1-pbinom(5,6,0.5)) [1] 0.03125
Since the result is significant at the 5% level of significance, there is some evidence from these data that the coin is biased.