Question

State College is evaluation a new English composition course for freshmen. A random sample of n = 25 freshmen is obtained and the students are placed in the course during their first semester. One year later, a writing sample is obtained for each student and the writing samples are graded using a standardized evaluation technique. The average score for the sample is M = 76. For the general population of college students, writing scores form a normal distribution with a mean of µ > 70. a. If the writing scores for the population have a standard deviation of σ = 20, does the sample provide enough evidence to conclude that the new composition course has a significant effect? Assume a two-tailed test with α = .05. b. If the population standard deviation is σ = 10, is the sample sufficient to demonstrate a significant effect? Again, assume a two-tailed test with α = .05. c. Comparing your answers for parts a and b, explain how the magnitude of the standard deviation influences the outcome of a hypothesis test.

The null hypothesis states that the population mean is (answer with whole number).    With σ = 20, the sample mean corresponds to z = (round answer to two decimal places). This (is OR is not) sufficient to reject the null hypothesis. You (can OR cannot) conclude that the course has a significant effect. However if the population standard deviation is σ = 10, the sample mean corresponds to z = (round answer to two decimal places). This (is OR is not) sufficient to reject the null hypothesis and conclude that the course (does OR does not) have a significant effect.

Answer 1

Solution a)
Null hypothesis: H0: μ = 70
Alternate Hypothesis: Ha: μ ≠ 70

Sample mean = 76
Standard deviation = 20
Standard error of mean = σ / √ n
Standard error of mean = 20 / √ 25
SE = 20/5
Standard error of mean 4
z = (xbar- μ ) / SE
z = (76-70) / 4
z = 1.5

P-value = P( |z| > 1.5) = 2 P( z > 1.5) = 2(0.0668) = 0.1336
P-value > 0.05, so don't reject H0

This (is not) sufficient to reject the null hypothesis. You (cannot) conclude that the course has a significant effect.

Solution b)
Sample mean = 76
Standard deviation = 10
Standard error of mean = σ / √ n
Standard error of mean = 10 / √ 25
SE = 10/5
Standard error of mean 2
z = (xbar- μ ) / SE
z = (76-70) / 2
z = 3

P-value = P( |z| > 3) = 2 P( z > 3) = (2)( 0.0013) = 0.0026
P-value < 0.05, so reject H0.
This (is) sufficient to reject the null hypothesis and conclude that the course (does) have a significant effect.

Solution c)
Smaller the standard deviation, greater the chance of the null hypothesis being rejected.