Question

There are about 18 million households in the US that self-identify as Hispanic and have annual income below $100 000. Aaron and Sophie disagree about what fraction of this population has average family income below about $50 000. Aaron thinks about 50% of this population has average family income below about $50 000. Sophie thinks Aaron’s fraction is just wrong. They agree to hypothesis test Aaron’s estimate at level of significance 0.01, using the following annual income estimates from a random sample of fifteen such families:
18500,78500,41000,48500,33500,46000,14000,56000,28500,31000,21000,36000,46000,16000,36000
The sample proportion is  
The value of the test statistic is  
The critical numbers are   
The P-value is

Answer 1

Number of people have salaries less than 50000 dollars is 2

So sample proportion is 2/15= 0.133

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA:

LEVEL OF SIGNIFICANCE=0.01

Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a left-tailed test is zc​=−2.33.

The rejection region for this left-tailed test is R={z:z<−2.33}

Test Statistics

The z-statistic is computed as follows:

Decision about the null hypothesis

Since it is observed that z=−2.84<zc​=−2.33, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0023, and since p=0.0023<0.01, it is concluded that the null hypothesis is rejected.

Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is less than 50% or 0.5, at the α=0.01 significance level.

So Aron is correct that 50% of this population has average family income below about $50 000.