**Question 2**

A professor wants to knowif her introductory statistics class has a good grasp of baic math. Six students are chosen at random form the calll an given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:

62, 92, 75, 68, 83, 95.

Can the professor have 90% confidence that the mean score for the class on the test would be above 70.

**Solution Steps**

I recommend you read the question a number of time and make sure you understand it very well and note the following:

- Variance is not given
- Sample size is only six
- The mean score that need to be tested is 70

**Quiz**: What test do we need to use here?**Step 1: Set up the null hypothesis and the alternate hypothesis**

**H**μ

_{0}:_{0}> 100

**H**: μ

_{a}_{0}<= 100

if the null hypothiesis requires that mean to be greater than 70, then for the alternate hypothesis, the mean should be less than or equal to 70.

**Step 2: Determine what type of test to use**

From the question, the sample size is very small, just 6 and the variance is not given. In this case we are going to use t-Test.(one-sample t-test)

**Step 3: Calculate the mean of the sample**

You can calculate the variance of the sample using the formular below. Same as formular for average

= (62 + 92 + 75 + 68 + 83 + 95) / 6 = 79.167

So we have a mean,

x̄

_{n}= 79.147

**Step 4: Calculate the variance of the sample**

You can calculate the variance of the sample using the formular below

This would give us a value of 173.38 (This is actually the variance

Taking the squareroot, would given us the standard deviation as 13.167

**Step 5: Calculate the tested statistics t**

The formular for the t-test is given below

This would give us

t = 1.705

**Step 6: Check value of t in the table**

Note the following

The question specifies 90% confidence.

Degrees of freedom is given as n - 1 = 6 -1 = 5

From the statistical tables, the critical value if t is given as 2.1328

**Step 7: Draw a conclusion**

Since the calculated value of t is less than the critical value of t, in this case, we accept the null hypothesis. So the professor can can be confident that the mean score of he class can be 70.