Question

ASSESSMENT

Perform as indicated

In numbers 1-5, find the critical valuets) and rejection remonts) for the type of z-test with level of significance a Include a graph with your answer

1 Left -tailed test, a=0.03

2 Right-tailed test = 0.05

3 Two-tailed test, a = 0.02

4 Two-tailed test, a = 0.10 5 Left-tailed test, a=0,09

In numbers 6-9, state whether each standardized test statistic z allows you to reject the null hypodesis Explain your reasoning

6. z=-1301 7 z=1203

8 2 1.280 9 2=1.286

10 A fast food restaurant estimates that the mean sodium content in one of its breakfast sandwiches is no

more than 920 milligrams A random sample of 44 breakfast sandwiches has a mean sodium content of 925 milligrams Aanume the population standard deviation is 15 milligrams At a=0.10, do you have enough evidence to reject the restaurant's claim?

Answer 1

1.

$$z\le-1.8808$$

2.

$$z\ge1.6449$$

3.

$$|z|\ge2.3263$$

4.

$$|z|\ge1.6449$$

5.

$$z\le -1.3408$$

Let we have left-tailed and $z_c=-1.2816$

6.

Reject $H_0$ because $z< −1.2816.$

7. Fail to reject $H_0$ because $z > −1.2816.$

8. Fail to reject $H_0$ because $z > −1.2816.$

9. Fail to reject $H_0$ because $z > −1.2816.$

10.

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le920$

$H_1:\mu>920$

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

Based on the information provided, the significance level is $\alpha = 0.10,$ and the critical value for a right-tailed test is $z_c = 1.2816.$

The rejection region for this right-tailed test is $R = \{z:z>1.2816\}.$

The z-statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{925-920}{15/\sqrt{44}}\approx2.211$$

Since it is observed that $z=2.211>1.2816=z_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value is $p=P(z>2.211)=0.027036,$ and since $p= 0.027036<0.10=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is greater than 920, at the $\alpha = 0.10$ significance level.