An educator is considering two different videotapes for use in a half-day session designed to introduce students to the basics of economics. Students have been randomly assigned to two groups, and they all take the same written examination after viewing the videotape. The scores are summarized here. Assuming normal populations with equal standard deviations, does it appear that the two videotapes cold be equally effective? What is the most accurate statement that could be made about the p-value for the test?
Videotape 1: ¯x1 = 77.1 s1 = 7.8 n1 = 25
Videotape 2: x¯2 = 80.0 s2 = 8.1 n2 = 25
First make a hypothesis, for example,
H : the two tapes are equally effective, or
H : &mu1 = &mu2
From the given data, calculate the t-statistic
t = (x̄2 - x̄1)/sqrt(s1²/n1+s2²/n2)
Calculate the degrees of freedom:
dof = ((s1²/n1) + (s2²/n2))²/((s1²/n1)²/n1 + (s2²/n2)²/n2)
Look up the Student's t-test values for the corresponding probability that they are equally effective. Use the two-tail test if there is no presumed domininance of one tape over the other.
A detailed numerical example is given in the following link:
http://www.jiskha.com/display.cgi?id=1249353992
To determine whether the two videotapes could be equally effective, we can conduct a two-sample t-test. The null hypothesis (H0) assumes that the means of the two populations are equal, while the alternative hypothesis (Ha) assumes that they are not equal.
Given the summary statistics:
Videotape 1: ¯x1 = 77.1, s1 = 7.8, n1 = 25
Videotape 2: x¯2 = 80.0, s2 = 8.1, n2 = 25
Now, let's calculate the test statistic (t):
t = (¯x1 - ¯x2) / √( ( s1^2 / n1 ) + ( s2^2 / n2 ) )
= (77.1 - 80.0) / √( ( (7.8)^2 / 25 ) + ( (8.1)^2 / 25 ) )
= -2.9 / √( 6.096 + 6.624 )
= -2.9 / √12.72
≈ -2.9 / 3.561
≈ -0.814
Next, we need to find the degrees of freedom (df), which is given by the formula:
df = ( ( s1^2 / n1 + s2^2 / n2 )^2 ) / ( ( ( ( ( s1^2 / n1 )^2 ) / ( n1 - 1 ) ) + ( ( ( s2^2 / n2 )^2 ) / ( n2 - 1 ) ) ) )
= ( ( ( (7.8)^2 / 25 ) + ( (8.1)^2 / 25 ) )^2 ) / ( ( ( ( ( (7.8)^2 / 25 )^2 ) / ( 25 - 1 ) ) + ( ( (8.1)^2 / 25 )^2 ) / ( 25 - 1 ) ) )
≈ ( ( 6.096 + 6.624 )^2 ) / ( ( ( ( ( ( 6.096^2 / 25 ) / 24 ) + ( 6.624^2 / 25 ) ) / 24 ) + ( 6.624^2 / 24 ) ) )
≈ ( 12.72^2 ) / ( ( ( ( ( 0.00509216 ) / 24 ) + ( 0.006936384 ) ) / 24 ) + ( 0.006936384 ) )
≈ 161.65408 / ( ( ( 0.00021217 ) + 0.006936384 ) / 24 + 0.006936384 )
≈ 161.65408 / ( 0.007148554 ) / 24
≈ 161.65408 / 0.171565583
≈ 942.305
Using these values, we can calculate the p-value associated with the test statistic using a t-distribution table or calculator.
The most accurate statement that could be made about the p-value for the test is that it is less than the significance level (α) chosen for the test. If α is not provided, it is commonly set to 0.05. By comparing the calculated p-value to the significance level, we can determine whether to reject or fail to reject the null hypothesis.
To determine whether the two videotapes could be equally effective in introducing students to the basics of economics, we can perform a hypothesis test for the means of two independent populations.
Let's set up the hypotheses:
Null hypothesis (H0): The mean scores for both videotapes are equal.
Alternative hypothesis (HA): The mean scores for the two videotapes are not equal.
To conduct the test, we can use the two-sample t-test since the population standard deviations are unknown.
Step 1: Calculate the test statistic
We need to calculate the t-value using the formula:
t = (x¯1 - x¯2) / sqrt((s1^2/n1) + (s2^2/n2))
Given the data:
¯x1 = 77.1, s1 = 7.8, n1 = 25 (for Videotape 1)
x¯2 = 80.0, s2 = 8.1, n2 = 25 (for Videotape 2)
t = (77.1 - 80.0) / sqrt((7.8^2/25) + (8.1^2/25))
Step 2: Calculate the degrees of freedom (df)
The degrees of freedom can be calculated using the formula:
df = (s1^2/n1 + s2^2/n2)^2 / ((s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1))
df = (7.8^2/25 + 8.1^2/25)^2 / ((7.8^2/25)^2/(25-1) + (8.1^2/25)^2/(25-1))
Step 3: Determine the p-value
Using the t-value and degrees of freedom, we can find the p-value associated with the test statistic.
The p-value is the probability of obtaining a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
Step 4: Make a decision
If the p-value is less than the significance level (α), typically 0.05, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Therefore, the most accurate statement about the p-value for the test is:
"If the p-value is less than 0.05, we reject the null hypothesis and conclude that the mean scores for the two videotapes are not equal. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis and conclude that there is insufficient evidence to suggest a difference in the mean scores."