Question

An engineer measured the diameter of bearings in cm.

  10.2 16.9 13.3 12.5 12.1 10.0 10.9 13.5 16.6 11.1 12.0 16.2 12.3 11.8 13.0

It is believed that the diameters of bearings form a normal distribution.

1. Find an efficient interval that 90% of the data would lie.
   
2. Find an efficient interval that the population mean diameter of bearings would lie with 90% confidence.
   
3. Test whether the population mean diameter of bearings is 14 at α =0.05 by the critical value approach.
   
4. Test if population mean diameter of bearings is less than 14 at α =0.1 by the p-value approach.
   (If there are the same steps or quantities as in (c) , you can skip them by simply referring to them or copying the result.)
   
5. Can we derived the conclusion of the test in (d) based on the interval derived in (b)?
   If yes, explain explicitly how we can draw the same conclusion based on the interval in (b). If not, explain explicitly why we can't.
   
6. Without any further computations, make a conclusion of test if the population mean diameter of bearings is 14 at  α =0.1.

Answer 1

Given,

$\small \bar{x}= 12.8267$

s=2.19

n = 15

Interval that the population mean diameter of bearings would lie with 90%,

We know that,

$\small Confidence\; Interval = \bar{x} \pm t * \frac{s} {\sqrt{n}}$

where t for 90% CI is t(14df) = 1.761

Therefore the 90% confidence interval is,

$\small Confidence\; Interval = [11.831 \; , \; 13.823]$

Test whether the population mean diameter of bearings is 14 at α =0.05 by the critical value approach

The hypothesis is,

$\small H_0 : \bar{x} = 14$

$\small H_1: \bar{x} \neq 14$

a = 0.05

We know that,

$\small t = \frac{\bar{x} - 14}{s/\sqrt n}$

$\small t = -2.0750$

The ctitical values are with degree of freedom (df = n - 1 = 14),

$\small t^* = \pm 2.1448$

Since the t value is within the range of critical value, we fail to reject the null hypothesis

Test if population mean diameter of bearings is less than 14 at α =0.1 by the p-value approach.

The hypothesis is,

$\small H_0 : \bar{x} = 14$

$\small H_1: \bar{x} < 14$

a = 0.1

We know that,

$\small t = \frac{\bar{x} - 14}{s/\sqrt n}$

$\small t = -2.0750$

The corresponding p value is 0.0285

Since the p value is less than the significance level(a = 0.01),, we reject the null hypothesis.

We can derive the conclusion from b, because we know tha 90% confidence the mean and we know that the interval has values lesser than 14. Therefore we can conclude the same as we have done for (d)