Question

I took a survey on 60 random people. 54 responded . Question was How many children do you have?

0 children-3 people responded

1 child-3 people responded

2 children-9 people responded

3 children-19 people

4-19 people responded

5+-2 people responded

I am needing to perform a hypothesis test for this question.

• • State whether the test is one- or two-tailed, state the decision rule that will determine significance or not
  • List all the variables for the equation including mean, null (your own expected value since you do not have population data for comparison), sample standard deviation, and sample size.
  • Write and solve the equation or excel command with values from your data.
  • Compare your test statistic with the decision rule and determine significance.
  • Make a one sentence statement about your decision regarding the null hypothesis.
• Update your conclusion. The updated conclusion must show reflection of the test hypothesis in final decisions.

Answer 1

I have only doubts about the null value in the Hypothesis test part. I shall make my own guess as suggested in your question.

Assumptions: The number of children follows a Normal distribution and are independent.

The null and alternate hypothesis:

(average number of children in the population is 3)

(I had taken one sided alternative)

(Note: I had taken two sided alternative since nothing is known about the population. Looking at the data, the frequency observed was more have number of children 3 and 4(38 persons) and hence, I took a two sided alternative).

Level of significance:

We have 55 persons responded and hence will assume n=55. The data is summarized and the mean and variance are calculated by following the steps below.

x	f	x*f	x2f
0	3	0	0
1	3	3	3
2	9	18	36
3	19	57	171
4	19	76	304
5	2	10	50
Total	55	164	564

We have

The mean number of children

The sample variance   

  

  

Therefore, for the given information we have

The test statistic: will follow a t distribution with n-1 df.

  

  

  

  

The region of rejection is .

Decision: Since the calculated value of falls in the rejection region, we reject the null hypothesis. hence we concluded that there is sufficient evidence to claim that the average number of children in the population is not equal to 3. The p-value for the test is p=0.9092.