Question

a) A machine is set to produce disc plates with a mean diameter of 14 mm. A sample of 8 discs gave a mean diameter, 𝑥̅ = 14.9 mm and a standard deviation, s = 1.33 mm. A test was carried out at the 5% level of significance to determine whether the machine is in good working order. Assume that the diameter of the disc follows a normal distribution. i. State, in symbols, the null and alternate hypotheses for this test. [2] ii. State, with reasons, whether a t-test or a z-test will be appropriate. [3] iii. (Determine the rejection region(s) of the test. [3] iv. Calculate the value of the test statistic. [3] v. State, with reason, a valid conclusion for the test. [2]

Answer 1

Method 1

$\bar{x} = 14.9 \\ s = 1.33 \\ n=8 \\ α=0.05$

$\mu=14mm.$

(i)

For the machine to be in a good working condition, the mean diameter of disc plates produced should be 14mm, otherwise the machine is in bad working condition.

Therefore,

$H_0:\mu=14$

$Against$

$H_1:\mu\not=14$

These are respectively the required null and alternative hypotheses in symbols to perform the test.

(ii) When the sample size n<30 and the population standard deviation $\sigma$ is unknown, a one sample t-test for the population mean $\mu$ is used, where the sample standard deviation $s$ is used.

(iii) Based on the information provided, the significance level is α=0.05 and df = n-1 = 7 degrees of freedom.

The critical value of the two-tailed test is $t_c = 2.365$

The rejection region is $R = \{t: |t| > 2.364619\}.$

(iv) The t-statistic is computed as follows:

$t = \frac{\bar{x} - \mu}{ s / \sqrt{n}} \\ t = \frac{14.9-14}{1.33 / \sqrt{8}} \\ t = 1.914$

(v) Using the critical value approach: since it is observed that $|t| = 1.913973 \le 2.364619= t_c ,$ it is then concluded that we fail to reject the null hypothesis. There is not enough evidence to claim that the population mean is different than 14 at α=0.05 significance level. Thus, the machine is in good working order.

Using the P-value approach: the p-value for two-tailed test with $\alpha=0.05, df=7$ degrees of freedom, $t=1.913973$ is $p = 0.097188,$ and since $p=0.097188\geq0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than 14 at the $\alpha = 0.05$ significance level.Thus, the machine is in good working order.

Method 2.

(i) Since the sample mean significantly greater than population mean, then we should make next hypothesis

$H_0:m=14$

$H_1:m>14$ , where m - population mean

(ii) Since we have small sample size, then it is appropriate to use t-test

(iii) According to the form of alternative hypothesis, right-tailed test is appropriate, then

$P(T(n-1)>Cr)=1-a$ , where T(n-1) - student's criteria with n-1 degrees of freedom, n - sample size, Cr - critical value a - level of significance

In the given case we have

$P(T(7)>Cr)=0.05\implies Cr=1.895$

If our t-statistic will be greater than 1.895, then we should reject null hypothesis in favor of alternative

(iv) test statistic calculated the following way

$t={\frac {x-a_0} s}*\sqrt{n}$

$t={\frac {x-a_0} s}*\sqrt{n}$ , where x - sample mean, $a_0$ - population mean according to the null hypothesis? s - sample standard deviation, n - sample size

In our case we have

$t={\frac {14.9-14} {1.33}}*\sqrt{8}=1.914$

$t={\frac {14.9-14} {1.33}}*\sqrt{8}=1.914$

(v) We receive that t > Cr, then, according to the given data, we have enough statistical evidence to reject the null hypothesis on 95% significance level and conclude that population mean is greater than 14 mm.