Question

1. The claim is that more than 75% of workers are satisfied with their job, and sample statistics include 600 employed adults; with 490 of them saying that they are satisfied with their job. Use a 0.01 confidence interval to test the claim.

Claim: Null Hypothesis: Alternative Hypothesis:

Test Statistic:

P-Value:

Interpreting the Test Statistic by using P-Value:

Critical Value:

Interpreting the Test Statistic by using Critical Value:

Conclusion:

Answer 1

The null and alternative hypothesis are:

• Null hypothesis, , i..e, the true proportion of workers are satisfied with their job is NOT DIFFERENT than 0.75.
• Alternative hypothesis, , i.e., the true proportion of workers are satisfied with their job is GREATER than 0.75.

At given signifiance level of we need to test the hypothesis.

Test statistic: The formula for the test statistic is given by

where,

the null hypothesized value for the population parameter p

the proportion in the sample of workers who satisfied with their job

sample size

Calculation for test statistic:

So, the test statistic is calculated as

P-value: Since we are testing a right tailed hypothesis, so the p-value is calculated as -

Decision: The significance level is and the p-value=0.00008

Since,

Conclusion: At significance level of sample data provides sufficient evidence to reject null hypothesis , hence we reject null hypothesis.

In other words, we conclude that "The true proportion of workers who are satisfied with their job is GREATER tham 75% or 0.75"

______________________________

Using critical value method:

For the given significance level the critical value is -

Decision: The test statistic is calculated above as and the critical value is

Since,

Conclusion: At significance level sample data provides sufficient evidene to reject null hypothesis , hence we reject null hypothesis and conclude that, "The true proportion of workers who satisfied with their job is GREATER than 75% or 0.75."