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In Europe the diameters of women's rings have mean 18.5 mm .Researchers claim that women in Jakarta have smaller fingers than women in Europe .The researchers took a random sample of 20 women in Jakarta and measured the diameters of their rings .The mean diameter was found to be 18.1 . Assuming that the diameters of women's rings in Jakarta have a normal distribution with standard deviation 1.1 carry out the hypothesis test at the 2.5 percent level to check whether the claim is justified .

I tried this on my own ,but cannot understand why root 20 must be used while standardising thenZ value . Please help .

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    $\begingroup$ For a $Z$-test $$Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$$ follows a standard normal distribution. We have $n=20$ in this case. $\endgroup$ – Remy Oct 20 at 6:22
  • $\begingroup$ Oh i haven't studied the root n part ,why do we need to divide by it , thanks ! $\endgroup$ – Sara Oct 20 at 6:24
  • $\begingroup$ So your question is regarding the derivation of this formula? $\endgroup$ – Remy Oct 20 at 6:44
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    $\begingroup$ $SD(X_i) = \sigma;\, Var(\sum_{i=1}^n X_i) = n\sigma^2;\, Var(\bar X) = \sigma^2/n;\, SD(\bar X) = \sigma/\sqrt{n}.$ $\endgroup$ – BruceET Oct 20 at 6:44
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One-sided z Test at 2.5% Level

You want to test $H_0: \mu_J \ge 18.5$ against $H_a: \mu_J < 18.5$ at level $\alpha = 0.025$ given the $\sigma_J = 1.2.$ With $n = 20$ normal observations we have $\bar X_J = 18.1.$

Because $\sigma_J$ is known, this is a z-test. Minitab statistical software has a procedure for this test. The output is shown below. You should verify that the standard error of the mean (SE Mean) $\sigma/\sqrt{n} = 0.246$ and that the statistic $Z = -1.63$ according to formulas in your book. You should also use printed tables of the standard normal CDF (or software) to verify the P-value 0.052.

Using a P-value to test: Even though $\bar X_J = 18.1 < \mu_J = 18.5$ the sample mean $18.1$ is not sufficiently smaller than the hypothetical population mean $18.5$ to say that the difference is statistically significant at the 2.5% level. This is because the P-value is not $\le 0.025.$

[Using a critical value to test: You can also find the 2.5% critical value $c = -1.96$ for this test. This is the value that cuts probability $0.025$ from the lower tail of a standard normal distribution. (Again here, you could use printed normal tables or software.) If you had $T \le c,$ then you would have been able to reject $H_0.$ Most statistical software uses P-values as a criterion for rejecting $H_0,$ as above. But you can also find and use the critical value $c$ to do the test.]

Minitab output:

One-Sample Z 

Test of μ = 18.5 vs < 18.5
The assumed standard deviation = 1.1

 N    Mean  SE Mean  97.5% Upper Bound      Z      P
20  18.100    0.246             18.582  -1.63  0.052

Graphical Display: The figure below (made using R) shows the standard normal density curve. The heavy black line shows the observed Z statistic; the P-value is represented by the area under the density curve to the left of this line. The dotted red line shows the critical value $c;$ the area under the density curve to the left of this line is 0.025.

enter image description here

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This is a known formula given in introductory statistics classes - it should be in your notes. This comes from the fact that if a random variable $X$ comes from a population that has a mean $\mu$ and standard deviation $\sigma$ and each $X_i$ are independent of each other then

$$\mathsf{Var}\left(\bar{X}\right)=\mathsf{Var}\left(\frac{1}{n}\sum X_i\right)=\frac{1}{n^2}\mathsf{Var}\left(\sum X_i\right)=\frac{1}{n^2}\sum\mathsf{Var}(X)=\frac{1}{n^2}\left(n\sigma^2\right)=\frac{\sigma^2}{n}$$

Here we make use of the fact that for a random variable $Y$, we have that $\mathsf{Var}(\alpha Y)=\alpha^2\mathsf{Var}(Y)$ and that when the $Y_i$'s are independently and identically distributed then the variance of the sum is the sum of the variances. Since the standard deviation is the square-root of the variance the standard deviation is

$$\sqrt{\frac{\sigma^2}{n}}=\frac{\sigma}{\sqrt{n}}$$

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