Question

The average of IQ test scores of 31 seventh-grade girls in Midwest school district is 89. IQ scores follow a Normal distribution with standard deviation 15. Treat these 31 girls as an SRS of all seventh-grade girls in this district. IQ scores in a board population are supposed to have mean 100. Is there evidence that the mean in this district differs from 100?

Answer 1

`p_v =2*P(t_(30)<-4.08)=0.00031`

If we compare the p value and a significance level given for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and the the actual true mean is significantly different from 100.

Explanation:

Data given and notation

`\bar X=89` represent the mean of seventh-grade girls in Midwest school district

`s=15` represent the standard deviation for the sample

`n=31` sample size

`\mu_o =100` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is different from 100, the system of hypothesis would be:

Null hypothesis:
`\mu = 100`

Alternative hypothesis:
`\mu \\eq 100`

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(89-100)/((15)/(√(31)))=-4.08`

Calculate the P-value

First we need to calculate the degrees of freedom given by:

`df=n-1=31-1=30`

Since is a two tailed test the p value would be:

`p_v =2*P(t_(30)<-4.08)=0.00031`

Conclusion

If we compare the p value and a significance level given for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we can reject the null hypothesis, and the the actual true mean is significantly different from 100.