Question

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random sample of subjects with high lead levels in their blood (sample 2). The statistics are summarized below. Use a 0.05 significance level to test the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ of people with high blood lead levels.

Low Blood Lead Level: n = 78 x = 92.88 s = 15.34
High Blood Lead Level: n = 21 x = 86.90 s = 8.99
a. State and label the null and alternative hypotheses.
b. State the value of the test statistic.
c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Answer 1

a. Null hypothesis:
`\mu_1 \leq \mu_2`

Alternative hypothesis:
`\mu_1 >\mu_2`

b.
`t=\frac{(92.88 -86.90)-(0)}{\sqrt{(15.34^2)/(78)}+(8.99^2)/(21)}=2.282`

c.
`p_v =P(t_(97)>2.287) =0.0122`

So with the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 \leq \mu_2`

Alternative hypothesis:
`\mu_1 >\mu_2`

Or equivalently:

Null hypothesis:
`\mu_1 - \mu_2 \leq 0`

Alternative hypothesis:
`\mu_1 -\mu_2>0`

Our notation on this case :

`n_1 =78` represent the sample size for group 1

`n_2 =21` represent the sample size for group 2

`\bar X_1 =92.88` represent the sample mean for the group 1

`\bar X_2 =86.90` represent the sample mean for the group 2

`s_1=15.34` represent the sample standard deviation for group 1

`s_2=8.99` represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

`t=\frac{(\bar X_1 -\bar X_2)-(\mu_(1)-\mu_2)}{\sqrt{(s^2_1)/(n_1)}+(s^2_2)/(n_2)}`

Where t follows a t distribution with
`n_1+n_2 -2` degrees of freedom. If we replace the values given we have:

`t=\frac{(92.88 -86.90)-(0)}{\sqrt{(15.34^2)/(78)}+(8.99^2)/(21)}=2.282`

Now we can calculate the degrees of freedom given by:

`df=78+21-2=97`

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

`p_v =P(t_(97)>2.287) =0.0122`

So with the p value obtained and using the significance level given
`\alpha=0.05` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).