Question

Determine if there is sufficient evidence to conclude the average amount of divorces is greater than or equal to 3000 in the United States and territories at the 0.05 level of significance N=52.

1. Clearly state a null and alternative hypothesis
2. Give the value of the test statistic
3. Report the P-Value
4. Clearly state your conclusion (Reject the Null or Fail to Reject the Null)
5. Explain what your conclusion means in context of the data.

Summary Table for Divorces
Mean	1,399
Median	1,176
Standard Deviation	1407
Minimum	0
Maximum	7,008

Answer 1

The provided sample mean is and the sample standard deviation is , and the sample size is n=52.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a left-tailed test is tc​=−1.675.

The rejection region for this left-tailed test is R=t:t<−1.675

(3) Test Statistics

The t-statistic is computed as follows

:

(4) Decision about the null hypothesis

Since it is observed that t=−8.205<tc​=−1.675, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is less than 3000, at the 0.05 significance level and consequently there is not sufficient evidence to conclude the average amount of divorces is greater than or equal to 3000.

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