Question

g A study was done to investigate the relationship of employment status to mental health. Each of the sampled unemployed workers was given a mental health examination using the General Health Questionnaire (GHQ). A GHQ score of more than 10 implies that the person is mentally unstable. For the 49 sampled unemployed workers, the mean and standard deviation of the GHQ scores were found to be 10.94 and 5.10, respectively. a) Do an appropriate hypothesis test to determine if unemployed workers tend to be mentally unstable. Use the significance level of α = 0.01. (4 Points) b) Calculate and interpret the p-value of the test. (3 Points)

Answer 1

a) Null hypothesis:
`\mu \leq 10`

Alternative hypothesis:
`\mu > 10`

We don't know the population deviation so then the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the data given we got:

`t=(10.94-10)/((5.10)/(√(49)))=1.29`

b)
`df=n-1=49-1=48`

`p_v =P(t_(48)>1.29)=0.102`

Since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the unemployed workers is significantly higher for the GHQ score

Explanation:

Information given

`\bar X=10.94` represent the sample mean for the GHQ score for unemployed workers

`s=5.10` represent the sample standard deviation

`n=49` sample size selected

`\mu_o =10` represent the limit value for the GHQ score

`\alpha=0.01` represent the significance level

t would represent the statistic

`p_v` represent the p value for the test

Part a : Hypothesis to check

We want to determine if unemployed workers tend to be mentally unstable (mean more than 10), the system of hypothesis would be:

Null hypothesis:
`\mu \leq 10`

Alternative hypothesis:
`\mu > 10`

We don't know the population deviation so then the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the data given we got:

`t=(10.94-10)/((5.10)/(√(49)))=1.29`

Part b: P value

The degrees of freedom for this test are:

`df=n-1=49-1=48`

Since the test is right tailed the p value would be:

`p_v =P(t_(48)>1.29)=0.102`

Since the p value is higher than the significance level we have enough evidence to conclude that the true mean for the unemployed workers is significantly higher for the GHQ score