Question

One hundred students were given an Algebra test. A random sample of ten students was taken out of class of 800 enrolled students. The time it took each student to complete the test is recorded below. 22.2, 23.7, 16.8, 18.3, 19.7, 16.9, 17.2, 18.5, 21.0, and 19.7

a. Find the mean, variance and standard deviation for this sample of ten students

b. Construct a 95% confidence interval for the population mean time to complete this Algebra test.

c. Test if the population mean time to complete the test is 22.5 minutes.

Answer 1

a)
`\bar X= (\sum_(i=1)^n X_i)/(10) =19.4`

`s^2 = (\sum_(i=1)^n (x_i -\bar x)^2)/(n-1)=5.438`

`s= √(5.438)=2.332`

b)
`19.4-2.262(2.332)/(√(10))=17.732`

`19.4+2.262(2.332)/(√(10))=21.068`

So on this case the 95% confidence interval would be given by (17.732;21.068)

c)
`t=(19.4-22.5)/((2.332)/(√(10)))=-4.203`

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

`p_v =2*P(t_(9)<-4.203)=0.00230`

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 22.5 minutes.

Explanation:

Part a

We have the following data given: 22.2, 23.7, 16.8, 18.3, 19.7, 16.9, 17.2, 18.5, 21.0, and 19.7

We can calculate the sample mean with this formula:

`\bar X= (\sum_(i=1)^n X_i)/(10) =19.4`

And the sample variance with this formula:

`s^2 = (\sum_(i=1)^n (x_i -\bar x)^2)/(n-1)=5.438`

And the sample deviation would be just the square root of the sample variance

`s= √(5.438)=2.332`

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X =19.4` represent the sample mean

`\mu` population mean (variable of interest)

s=2.332 represent the sample standard deviation

n=10 represent the sample size

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=10-1=9`

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(0.025,9)".And we see that
`t_(\alpha/2)=2.262`

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

`19.4-2.262(2.332)/(√(10))=17.732`

`19.4+2.262(2.332)/(√(10))=21.068`

So on this case the 95% confidence interval would be given by (17.732;21.068)

Part c

Null hypothesis:
`\mu =22.5`

Alternative hypothesis:
`\mu \\eq 22.5`

Since we don't know the population deviation, 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=(19.4-22.5)/((2.332)/(√(10)))=-4.203`

P-value

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

`p_v =2*P(t_(9)<-4.203)=0.00230`

Conclusion

If we compare the p value and the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly different from 22.5 minutes.