3 Hypothesis Testing

3.3 Concrete items

A construction company receives concrete items for a construction. The length of the items are assumed reasonably normally distributed. The following requirements for the length of the elements are made $\mu=3000mm$ . The company samples 9 items from a delivery which are then measured for control. The following measurements (in mm) are found: 3003, 3005, 2997, 3006, 2999, 2998, 3007, 3005, 3001.

a)

To investigate whether the requirement to the mean is fulfilled (with $\alpha= 5%$ ), the following hypothesis should be tested

$H_{0}:\ =\ 3000 \\ H_{1}:\ \neq\ 3000$

Or similarly asked: what is the evidence against the null hypothesis?

We know that:

$\bar{x} = 3002.333 \\ s = 3.708099 \\$

By using Method 3.36

1.
$$ t_{obs}= \frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}=1.88749 $$

> (3002.333-3000)/(3.708099/sqrt(9))
[1] 1.88749

2. $$ \text{p-value}= 2*P(T>|t_{obs}|) = 0.09579526 $$

> 2*(1-pt(1.88749, df=8))
[1] 0.09579526

3. Conclusion:

$\text{p-value}$ (0.096) is more than $\alpha$ (0.05), so we can accept $H_{0}$

Note that we can produce same result by using t.test in R

> t.test(sample, mu=3000)

    One Sample t-test

data:  sample
t = 1.8878, df = 8, p-value = 0.09576
alternative hypothesis: true mean is not equal to 3000
95 percent confidence interval:
 2999.483 3005.184
sample estimates:
mean of x 
 3002.333

b)

What would the level $\alpha = 0.01 $ critical values be for this test, and what are the interpretation of these?

$t_{\alpha/2} = t_{0.005}= -3.355387 \\ t_{1-\alpha/2}= t_{0.995} = 3.355387$

> qt(0.005, df=8)
[1] -3.355387
> qt(0.995, df=8)
[1] 3.355387

We can reject $H_0$ if the observed test-statistics ( $t_{obs}$ ) is more extreme than the calculated critical value with the significance level of $\alpha=0.01$

c)

What would the level $\alpha=0.05$ critical values be for this test (compare also with the values found in the previous question)?

$t_{\alpha/2} = t_{0.025}= -2.306004 \\ t_{1-\alpha/2}= t_{0.975} = 2.306004$

> qt(0.05/2, df=8)
[1] -2.306004
> qt(1-0.05/2, df=8)
[1] 2.306004

Critical values get closer to each other as the $\alpha$ increases.

d)

Investigate, by some plots, whether the data here appears to be coming from a normal distribution (as assumed until now)?

hist(sample, main = "Histogram of Concrete items", xlab= "Measurements")

qqnorm(sample, ylab="Measurements Quantiles")
qqline(sample)

qqwrap <- function(x, y, ...){
  stdy <- (y-mean(y))/sd(y)
  qqnorm(stdy, main="", ...)
  qqline(stdy)}

wallyplot(x-mean(sample), FUN=qqwrap, ylim=c(-3,3))

e)

Assuming that you, maybe among different plots, also did the normal q-q plot above, the question is now: What exactly is plotted in that plot? Or more specifically: what are the x- and y-coordinates of e.g. the two points to the lower left in this plot?

X coordinates represent the expected normal quantiles.
Y coordinates represent the sample quantiles.

For example, the point $( 0, 3002.333 )$ represents respectively the 0.5 quantiles of the expected normal distribution and the sample.

3.4 Aluminum profile

The length of an aluminum profile is checked by taking a sample of 16 items whose length is measured. The measurement results from this sample are listed below, all measurements are in mm: 180.02, 180.00, 180.01, 179.97, 179.92, 180.05, 179.94, 180.10,180.24, 180.12, 180.13, 180.22, 179.96, 180.10, 179.96, 180.06 . From data is obtained: $\bar{x}=180.05 $ and $s = 0.0959$ . It can be assumed that the sample comes from a population which is normal distributed.

a)

Find the evidence against the following hypothesis:

$H_{0}\ : =\ \mu=180$

> t.test(aluminum, mu=180)

    One Sample t-test

data:  aluminum
t = 2.0851, df = 15, p-value = 0.05456
alternative hypothesis: true mean is not equal to 180
95 percent confidence interval:
 179.9989 180.1011
sample estimates:
mean of x 
   180.05

$\text{p-value}$ (0.05452038) is more than 0.05, but less than 1, so there is a weak evidences against $H_{0}$ .

b)

If the following hypothesis test is carried out

$H_{0}:\ \mu=180 \\ H_{1}:\ \mu\neq180$

What are the level $\alpha= 1%$ critical values for this test?

$t_{\alpha/2} = t_{0.005}= -2.946713 \\ t_{1-\alpha/2}= t_{0.995} = 2.946713$

> qt(0.005, df=16-1)
[1] -2.946713
> qt(0.995, df=16-1)
[1] 2.946713

c)

What is the 99%-confidence interval for $\mu$ ?

$\bar{x}\pm t_{0.995}* \frac{s}{\sqrt{n}} = 180.05 \pm 2.946713*\frac{0.0959}{\sqrt{16}} = \\ = 180.05 \pm 0.07064744 = [ 179.9794, 180.1206]$

> qt(0.995,15)
[1] 2.946713

d)

Carry out the following hypothesis test using $\alpha=5\%$

$H_{0}:\ \mu=180 \\ H_{1}:\ \mu\neq180$

We know that:

$\bar{x} = 180.05 \\ s = 0.0959 \\$

By using Method 3.36

1.
$$ t_{obs}= \frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}= 2.085506 $$

> (180.05-180)/(0.0959/sqrt(16))
[1] 2.085506

2. $$ \text{p-value}= 2*P(T>|t_{obs}|) = 0.05452038 $$

> 2*(1-pt(2.085506, df=15))
[1] 0.05452038

3. Conclusion:

$\text{p-value}$ (0.05452038) is more than $\alpha$ (0.05), so we can accept $H_{0}$ .

> t.test(aluminum, mu=180)

    One Sample t-test

data:  aluminum
t = 2.0851, df = 15, p-value = 0.05456
alternative hypothesis: true mean is not equal to 180
95 percent confidence interval:
 179.9989 180.1011
sample estimates:
mean of x 
   180.05

$\text{p-value}$ (0.05452038) is more than 0.05, but less than 1, so there is weak evidences against $H_{0}$ .