Question

Microsoft excel was used on a set of data involving the number of defective items found in a random sample of 46 cases of light bulbs produced during a morning shift at a plant. A manager wants to know if the mean number of defective bulbs per case is greater than 20 during the morning shift. She will make her decision using a test with a level of significance of 0.10. The following information was extracted from the microsoft excel output for the sample of 46 cases:
n=46, Arithmetic mean=28.00, Std Dev =25.92, standard error=3.82, Null hypothesis: H0 : u<=20, alpha =0.10, df=45, t-test statistic=2.09, one tail test upper critical value =1.3006, p-value=0.021
i) what parameter is the manager interested in?
ii) state the alternative hypothesis for this study.
iii) what critical value should the manager use to determine the rejection region.
iv) explain if the, null hypothesis should be rejected and why or why not?
v) explain our risk of committing of a type1 error.
vi) explain if the data evidence proves beyond a doubt that the mean number of defective bulbs per case is greater than 20 during the morning shift
vii) what can the manager conclude about the mean number of defective bulbs per case during the morning shift using a level of significance of 0.10?
viii) what would the p-value be if these data were used to perform a two tail test?

Answer 1

Answer:

Step-by-step explanation:

i. The parameter the manager is interested in is number of defective bulbs in a case.

ii. Null hypothesis: u <= 20

Alternative hypothesis: u > 20

iii. The critical value the manager should use to determine the rejection region is 1.645.

iv. Using the p value which is 0.021 at 0.10 significance level we will reject the null as the p value is less than 0.1. Thus, we will conclude that there is enough statistical evidence to prove that the mean number of defective bulbs per case is greater than 20.

v. Our risk of committing type one error is alpha which is the level of significance set for the hypothesis test. An alpha level of 0.1 shows that we are willing to accept a 10% chance that we are wrong when you reject the null hypothesis.

vi. With a low p value, the data has enough evidence to prove that the mean number of defective bulbs per case is greater than 20 during the morning shift

vii. The manager will conclude that there is sufficient statistical evidence to prove that mean number of defective bulbs per case is greater than 20 during the morning shift.

viii. the p value if this is a two tail test would be 0.03662