Question

The calibration of a scale is to be checked by weighing a 10-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 = .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 = 9.85, and what would you conclude at significance level .01? (c) For a test with α = .01, what is the probability that recalibration is judged unnecessary when in fact µ = 10.2?

Answer 1

Answer:

a:  That the mean weight of the trials is 10 kg

b:  See attached photo for work

Step-by-step explanation:

We want to see if the scale is weighing properly and are using a 10 kg weight to calibrate it.  That means our hypothesis test is that the mean weight of the trails (in this case 25) is 10 kg.

The hypothesis we will use are

H0:  µ = 10

Ha:  µ ≠ 10

The alternate hypothesis has a not equals to sign because if the scale weighs too much or to little, then it needs to be better calibrated, so it's a two tailed test.

Answer 2

The calibration of the scale follows a normal distribution.

• The null and the alternate hypotheses are: $\mathbf{H_o: \mu = 10}$ and $\mathbf{H_a: \mu \ne 10}$
• The p-value when $\mathbf{\bar x = 9.85}$ is $\mathbf{p=0.000494}$
• The scale needs to be calibrated
• The probability that  recalibration is judged unnecessary is less than 0.00001

The given parameters are:

$\mathbf{\sigma = 0.200}$

$\mathbf{\mu = 10}$

$\mathbf{n = 25}$

$\mathbf{\bar x = 9.85}$

(a) The null and the alternate hypotheses

The true average weight is to be tested.

So, the null and the alternate hypotheses are:

$\mathbf{H_o: \mu = 10}$

$\mathbf{H_a: \mu \ne 10}$

(b) The p-value when $\mathbf{\bar x = 9.85}$

First, we calculate the test statistic

$\mathbf{t = \frac{\bar x - \mu}{\sigma/\sqrt n}}$

So, we have:

$\mathbf{t = \frac{9.85 - 10}{0.2/\sqrt{25}}}$

$\mathbf{t = \frac{9.85 - 10}{0.2/5}}$

$\mathbf{t = \frac{-0.15}{0.04}}$

$\mathbf{t = -3.75}$

Using p-value calculator, we have:

$\mathbf{p=0.000494}$

The critical regions of $\mathbf{t = -3.75}$ are t >2.797 and t < -2.797

Because -3.75 < -2.797, we reject the null hypothesis.

This means that, the scale needs to be calibrated

(c) Probability that recalibration is judged when $\mathbf{\mu = 10.2}$

First, we calculate the test statistic

$\mathbf{t = \frac{\bar x - \mu}{\sigma/\sqrt n}}$

So, we have:

$\mathbf{t = \frac{9.85 - 10.2}{0.2/\sqrt{25}}}$

$\mathbf{t = \frac{-0.35}{0.04}}$

$\mathbf{t = -8.75}$

Using p-value calculator, we have:

$\mathbf{p<0.00001}$

The probability that  recalibration is judged unnecessary is less than 0.00001

Read more about probabilities using test statistics at:

https://brainly.com/question/22783864