A Statistics professor has observed that for several years students score an average of 106 points out of 150 on the semester exam. A salesman suggests that he try a statistics software package that gets students more involved with computers, predicting that it will increase students' scores. The software is expensive, and the salesman offers to let the professor use it for a semester to see if the scores on the final exam increase significantly. The professor will have to pay for the software only if he chooses to continue using it. In the trial course that used this software, 241 students scored an average of 109 points on the final with a standard deviation of 8.3 points. Complete parts a) and b) below.

What is test statistics? (t= )

Does this improvement seem to be practicallysignificant?

Determine the 95% confidence interval for the mean score using the software, rounding to one decimal place.

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 106

For the alternative hypothesis,

µ > 106

It is a right tailed test because of >.

Since no population standard deviation is given, the distribution is a student's t.

Since n = 241,

Degrees of freedom, df = n - 1 = 241 - 1 = 240

t = (x - µ)/(s/√n)

Where

x = sample mean = 109

µ = population mean = 106

s = samples standard deviation = 8.3

a) test statistic, t = (109 - 106)/(8.3/√241) = 5.61

We would determine the p value using the t test calculator. It becomes

p = 0.00001

Since alpha, 0.05 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 5% level of significance, this improvement seem to be practicallysignificant.

b) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

Where

s = sample standard deviation

n = number of samples

From the information given, the population standard deviation is unknown, hence, we would use the t distribution to find the z score

df = 240

Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 1.6512

Margin of error = 1.6512 × 8.3/√241

= 0.88

Confidence interval = 109 ± 0.88

eve and marks ratios are equivalent

step-by-step explanation: