Question

The calibration of a scale is to be checked by weighing a 11 kg test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with σ = 0.200 kg. Let μ denote the true average weight reading on the scale.

a. What hypotheses should be tested?
b. With the sample mean itself as the test statistic, what is the P-value when x = 10.82?(Round your answer to four decimal places.)
c. What would you conclude at significance level 0.01?
d. For a test with α = 0.01,what is the probability that recalibration is judged unnecessary when in fact μ = 11.1?(Round your answer to four decimal places.)
e. For a test with α = 0.01,what is the probability that recalibration is judged unnecessary when in fact μ = 10.9? (Round your answer to four decimal places.)

Answer 1

Answer:

a) The hypothesis that should be tested are:

$H_0: \mu=11\\\\H_a:\mu\neq 11$

b) P-value = 0.00001

c) There is  enough evidence to support the claim that the scale measure differs significantly from 11.

d) P=0.5303

e) P=0.5303

Step-by-step explanation:

a. The hypothesis will depend on the claim that want to be tested. In this case, we want to know if the scale si correctly calibrated. Then, the alternative  hypothesis will be that the weight measured by the scale differs from 11 kg (in another words, the scale is not well calibrated).

This would be tested again a null hypothesis, that claims that there is no significative difference from 11 kg and the differences are due to pure chance.

Then, the hypothesis can be written as:

$H_0: \mu=11\\\\H_a:\mu\neq 11$

b) The sample has a size n=25.

The sample mean is M=10.82.

The standard deviation of the population is known and has a value of σ=0.2.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.2}{\sqrt{25}}=0.04$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{10.82-11}{0.04}=\dfrac{-0.18}{0.04}=-4.5$

This test is a two-tailed test, so the P-value for this test is calculated as:

$P-value=2\cdot P(z<-4.5)=0.00001$

c) As the P-value (0.00001) is smaller than the significance level (0.01), the effect is  significant.

The null hypothesis is rejected.

There is  enough evidence to support the claim that the scale measure differs significantly from 11.

d) If the population mean is μ=11.1, but we are testing against a null hypothesis that states  μ=11, to calculate the probability that recalibration is judged unnecessary, we have to calculate the critical values for μ=11 and calculate the probabilities for the acceptance region with the real mean.

The critical values for a significance level of 0.01 and a two-tailed sample is z=±2.576.

For a distribution N(11,0.2^2), this values correspond to the critical valules:

$X_1=\mu+z_1\cdot\sigma/\sqrt{n}=11+2.576\cdot 0.2\sqrt{25}=11+0.10304=11.1030\\\\X_2=\mu+z_2\cdot\sigma/\sqrt{n}=11-2.576\cdot 0.2\sqrt{25}=11-0.10304=10.8970$

To calculate the the probability of accepting a false null hypothesis, we have to recalculate the z-scores for the real mean:

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{11.1030-11.1}{0.2/\sqrt{25}}=\dfrac{0.0030}{0.04}=0.076 \\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{10.8970-11.1}{0.2/\sqrt{25}}=\dfrac{-0.2030}{0.04}=-5.076$

We will accept the null hypothesis (and not recomend the recalibration) with a probability calculated as:

$P=P(z_2<z<z_1)=P(z<z_1)-P(z<z_2)\\\\P=P(z<0.076)-P(z<-5.076)\\\\P=0.5303-0=0.5303$

e) In this case, we recalculate for a true mean of μ = 10.9.

$z_1=\dfrac{X_1-\mu}{\sigma/\sqrt{n}}=\dfrac{10.89696-10.9}{0.2/\sqrt{25}}=\dfrac{-0.003}{0.04}=-0.076\\\\\\z_2=\dfrac{X_2-\mu}{\sigma/\sqrt{n}}=\dfrac{11.10304-10.9}{0.2/\sqrt{25}}=\dfrac{0.203}{0.04}=5.076$

We will accept the null hypothesis (and not recomend the recalibration) with a probability calculated as:

$P=P(z_2<z<z_1)=P(z<z_1)-P(z<z_2)\\\\P=P(z<5.076)-P(z<-0.076)\\\\P=1-0.4697=0.5303$