Question

The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms. Statistical analysis of the output suggests that the resistances can be approximated by a normal distribution with known standard deviation of 0.005 ohms. We are interested in testing the hypothesis that the resistors conform to the specifications.

Requied:
a. Determine whether a random sample of 10 resistors yielding a sample mean of 0.152 ohms indicates that the resistors are conforming. Use alpha = 0.05.
b. Calculate a 95% confidence interval for the average resistance. How does this interval relate to your solution of part (a)?

Answer 1

a) The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.

b) The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.

Explanation:

Question a:

The resistors produced by a manufacturer are required to have an average resistance of 0.150 ohms.

At the null hypothesis, we test if this is the average resistance, that is:

`H_0: \mu = 0.15`

We are interested in testing the hypothesis that the resistors conform to the specifications.

At the alternative hypothesis, we test if it is not conforming, that is, the mean is different of 0.15, so:

`H_1: \mu \\eq 0.15`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

0.15 is tested at the null hypothesis:

This means that
`\mu = 0.15`

Sample mean of 0.152, sample of 10, population standard deviation of 0.005.

This means that
`X = 0.152, n = 10, \sigma = 0.005`

Value of the test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.152 - 0.15)/((0.005)/(√(10)))`

`z = 1.26`

P-value of the test and decision:

The p-value of the test is the probability of the sample mean differing from 0.15 by at least 0.152 - 0.15 = 0.002, which is P(|z| > 1.26), given by two multiplied by the p-value of z = -1.26.

Looking at the z-table, z = -1.26 has a p-value of 0.1038.

2*0.1038 = 0.2076

The p-value of the test is 0.2076 > 0.05, which means that the sample indicates that the resistors are conforming.

Question b:

We have to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a p-value of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(0.005)/(√(10)) = 0.003`

The lower end of the interval is the sample mean subtracted by M. So it is 0.15 - 0.003 = 0.147.

The upper end of the interval is the sample mean added to M. So it is 0.15 + 0.003 = 0.153.

The 95% confidence interval for the average resistance is (0.147, 0.153). 0.152 is part of the confidence interval, which means that as the test statistic in item a), it indicates that the resistors are conforming.