Question

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical population standard deviation s=180 is assumed known each year, the assistant dean uses a sample of applications to determine whether the mean examination score for the new freshman applications has changed.

- State the hypotheses

- What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean of Xbar=935 ?

- Use the confidence interval to conduct a hypthesis test. Using a=.05, what is your conclusion ?

- What is the p-value ?

Answer 1

a) Null hypothesis:
`\mu = 900`

Alternative hypothesis:
`\mu \\eq 900`

b) The 95% confidence interval would be given by (909.926;960.074)

c) Since our interval not contains the value of 900 we can reject the null hypothesis that the mean is equal to 900

d)
`p_v =2*P(t_((199))>2.75)=0.0065`

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is different from 900 at 5% of signficance.

Explanation:

Data given and notation

`\bar X=935` represent the sample mean

`s=180` represent the sample standard deviation

`n=200` sample size

`\mu_o =900` represent the value that we want to test

`\alpha=0.05` 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 check if the mean is equal to 900 or not, the system of hypothesis would be:

Null hypothesis:
`\mu = 900`

Alternative hypothesis:
`\mu \\eq 900`

Confidence interval

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=200-1=199`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,199)".And we see that
`t_(\alpha/2)=1.97`

Now we have everything in order to replace into formula (1):

`935-1.97(180)/(√(200))=909.926`

`935+1.97(180)/(√(200))=960.074`

So on this case the 95% confidence interval would be given by (909.926;960.074)

Part c

Since our interval not contains the value of 900 we can reject the null hypothesis that the mean is equal to 900

Part d

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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)))` (2)

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 (2) the info given like this:

`t=(935-900)/((180)/(√(200)))=2.75`

P-value

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

`p_v =2*P(t_((199))>2.75)=0.0065`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is different from 900 at 5% of signficance.