10.51 A statistics professor compared the scores of her morning class and afternoon class on the same
quiz. The 42 morning students’ average score was 83.52 with a standard deviation of 7.81, while
the 46 afternoon students’ average score was 79.44 with a standard deviation of 6.62. (a) Which
test would you use to compare the means? (b) State the hypotheses to see if there was a difference.
(c) Obtain a test statistic and p-value. (d) Why might there be a difference between morning and
afternoon scores? Explain. (e) Is it reasonable to assume equal variances? (f) Carry out a formal
test for equal variances at á = .05, showing all steps clearly.
could you please help explain the steps to this problem
Try a two-sample test.
Hypothesis:
Ho: µ1 = µ2
Ha: µ1 does not equal µ2
This is a two-tailed test because the problem is just asking if there is a difference (no specific direction in the alternate hypothesis).
P-value is the actual level of the test statistic (using a z or t-table, depending on the test you use).
A test involving two variances (standard deviation is the square root of the variance) would include these hypotheses: Ho would be the ratio of the two variances less than or equal to 1 and Ha would be the ratio of the two variances greater than 1.
Test statistic = (7.81)^2/(6.62)^2 = ?
Note: ^2 means squared.
Finish the calculation. Using an F-distribution table at .05 level of significance, determine the critical or cutoff value to reject the null. (You will need to use degrees of freedom to determine the value.) If the test statistic exceeds the critical value from the table, the null is rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null is not rejected.
I hope these few hints will help get you started.
Sure! Let's go through the steps of this problem one by one:
(a) Which test would you use to compare the means?
Since we are comparing the means of two independent samples, we can use the independent two-sample t-test.
(b) State the hypotheses to see if there was a difference.
The null hypothesis (H0) assumes that there is no difference between the two sample means, while the alternative hypothesis (Ha) assumes that there is a significant difference between the two sample means. So, the hypotheses can be stated as follows:
H0: µ1 = µ2 (There is no difference between the means of morning and afternoon class scores.)
Ha: µ1 ≠ µ2 (There is a significant difference between the means of morning and afternoon class scores.)
(c) Obtain a test statistic and p-value.
To obtain the test statistic and p-value, we can use the formulas for the independent two-sample t-test.
Test Statistic:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
Where,
x1 and x2 are the sample means,
s1 and s2 are the sample standard deviations,
n1 and n2 are the sample sizes.
In this case:
x1 = 83.52, x2 = 79.44
s1 = 7.81, s2 = 6.62
n1 = 42, n2 = 46
Plug in the values and calculate:
t = (83.52 - 79.44) / sqrt((7.81^2 / 42) + (6.62^2 / 46))
Calculate this value to get the test statistic.
To find the p-value, we need to consult the t-distribution table or use statistical software.
(d) Why might there be a difference between morning and afternoon scores? Explain.
There could be various reasons for the difference between morning and afternoon scores. For example, factors like differing levels of alertness, attention, or fatigue could impact students' performance. Additionally, different teaching styles, classroom dynamics, or even the time of day itself can influence student performance.
(e) Is it reasonable to assume equal variances?
To determine if it is reasonable to assume equal variances, we can perform a formal test for equal variances, such as the Levene's test or the F-test.
(f) Carry out a formal test for equal variances at α = .05, showing all steps clearly.
To perform a formal test for equal variances, we can use the F-test. The null hypothesis (H0) assumes equal variances, while the alternative hypothesis (Ha) assumes unequal variances.
The test statistic for the F-test is given by:
F = s1^2 / s2^2
Where,
s1 and s2 are the sample standard deviations.
In this case:
s1 = 7.81, s2 = 6.62
Calculate this value to get the test statistic for equal variances.
To obtain the p-value for the F-test, we can consult the F-distribution table or use statistical software.
I hope this helps! Let me know if you have any further questions.
Certainly! Here are the steps to follow for this problem:
(a) To compare the means of two groups, you would use a t-test for independent samples. This is because we are comparing the scores of two different classes, and each class is considered an independent sample.
(b) The hypotheses to test for a difference between the means are:
Null hypothesis (H0): There is no difference between the average scores of the morning class and the afternoon class. (µ1 = µ2)
Alternative hypothesis (Ha): There is a difference between the average scores of the morning class and the afternoon class. (µ1 ≠ µ2)
Where µ1 represents the mean score of the morning class and µ2 represents the mean score of the afternoon class.
(c) To obtain a test statistic and p-value, you can follow these steps:
1. Calculate the pooled standard deviation (Sp) using the formula:
Sp = sqrt(((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2))
Where n1 and n2 are the sample sizes of the morning and afternoon classes, and s1 and s2 are their respective standard deviations.
2. Calculate the t-statistic using the formula:
t = (x̄1 - x̄2) / sqrt((Sp^2/n1) + (Sp^2/n2))
Where x̄1 and x̄2 are the sample means of the morning and afternoon classes respectively.
3. Determine the degrees of freedom (df) using the formula:
df = n1 + n2 - 2
4. Use the t-distribution with the calculated df to find the p-value corresponding to the obtained t-statistic.
(d) There might be a difference between morning and afternoon scores due to various reasons like differing levels of alertness, focus, or effectiveness of teaching during different parts of the day, etc. However, the specific reason can only be determined by further investigation or analysis.
(e) To determine if it is reasonable to assume equal variances, you can conduct a formal test of equal variances. This is commonly known as Levene's test.
(f) To conduct a formal test for equal variances at α = 0.05, the steps are as follows:
1. Set up the hypotheses:
Null hypothesis (H0): The variances of the morning and afternoon classes are equal. (σ1^2 = σ2^2)
Alternative hypothesis (Ha): The variances of the morning and afternoon classes are not equal. (σ1^2 ≠ σ2^2)
2. Calculate the test statistic using the formula:
W = (n1 + n2 - 2) * (s1^2 + s2^2) / ((n1 + n2 - 4) * s1^2 + (n1 + n2 - 4) * s2^2)
Where n1 and n2 are the sample sizes of the morning and afternoon classes, and s1 and s2 are their respective standard deviations.
3. Determine the degrees of freedom for the test statistic, which is (n1 + n2 - 2).
4. Compare the test statistic to the critical value from the F-distribution table at α = 0.05. If the test statistic is larger than the critical value, reject the null hypothesis and conclude that the variances are not equal. If the test statistic is smaller than or equal to the critical value, fail to reject the null hypothesis and conclude that the variances are equal.
I hope this explanation helps you navigate through the steps of this problem. Let me know if you have any further questions!