Answer to Question #196109 in Statistics and Probability for Betty

Question

A renowned steel company produces metal products with a mean weight of 118 kg and standard deviation of 30 kg. Suppose a sample of 50 products is randomly taken and its mean weight was found to be 120 kg.

a) State the appropriate hypotheses to test whether the mean of sample products differ from the given mean weight.

b) Describe Type I and Type II errors in this context, and comment on the severity of each type of error.

c) Test the hypotheses stated in part (a) at a 1% level of significance, and comment on your findings.

Answer 1

a)

Hypothesized Population Mean $\mu=118$ 

Population Standard Deviation $\sigma=30$

Sample Size $n=50$

Sample Mean $\bar{X}=120$

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0:\mu=118$

$H_1:\mu\not=118$

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

b)

In a hypothesis tests there are two types of errors.

Type I error occurs when we reject a true null hypothesis.

Type II error occurs when we fail to reject a false null hypothesis.

The value of $\alpha$ , called the significance level of the test, represents the probability of making aType I error. In other words, $\alpha$ is the probability of rejecting the null hypothesis, $H_0,$ when in fact it is true.

$\alpha=P(H_0\ \text{is rejected}|H_0 \text{is true})$

The value of $\beta$ represents the probability of committing a Type II error; that is,

$\beta=P(H_0\ \text{is not rejected}|H_0 \text{is false})$

The value of $1-\beta$ is called the power of the test. It represents the probability of not making a Type II error.

c)

Significance Level $\alpha=0.01$ 

Based on the information provided, the significance level is $\alpha=0.01,$ and the critical value for a two-tailed test is $z_c=2.5758.$ 

The rejection region for this left-tailed test is $R=\{z:|z|>2.5758\}.$ 

The z-statistic is computed as follows:

$z=\dfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}}=\dfrac{120-118}{30/\sqrt{50}}$ $=\dfrac{\sqrt{2}}{3}\approx0.4714$

Since it is observed that $|z|=0.4714<2.5758=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $118,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value is $p=0.637355,$ and since $p=0.637355>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $118,$ at the $\alpha=0.01$ significance level.