2. The following is a random sample of grades achieved on statistics examination by 10 male students in a very large class and a random sample of grades achieved by 8 female students.
Male 78 87 75 64 90 80 65 86 74 81
Female 71 50 80 75 90 61 85 81
Test at = 0.01 level to test whether the difference between the means of these two samples is significant.
Generate an SPSS output and write the 5-step solutions for each of the following problems. All answers should be in two-decimal places:
1. A non-government consumer protection agency wants to test the credibility of the manufacturer's claim that the average capacity is 140 ampere-hours. To test the claim, 20 batteries from a recently produced batch were tested. The results, in ampere-hours are as follows:
140.0 137.5 138.9 140.0 144.0 139.1
141.9 137.4 133.6 138.2 140.1 139.6
136.8 137.3 35.0 139.0 141.3 140.6
135.3 134.2 Test at = 0.05
2. The following is a random sample of grades achieved on statistics examination by 10 male students in a very large class and a random sample of grades achieved by 8 female students.
Male 78 87 75 64 90 80 65 86 74 81
Female 71 50 80 75 90 61 85 81
Test at = 0.01 level to test whether the difference between the means of these two samples is significant.
3. To reduce the number of plant accidents. BLBTZ Co. implemented a new safety program for employees. To determine its effectiveness the number of accidents a week was recorded before and after the program was implemented.
Week 1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th
Before 12 10 9 15 11 8 16 13 14 20
After 10 9 12 8 6 10 12 7 5 11
Test at - 0.05 if the new safety program is effective in reducing accidents.
4. The following are the number of minutes it took a random sample of 10 men and 12 women applicants to complete applications for employment at the human resource department of ABC Corporation.
Men: 19.4 18.5 19.8 18.9 21.0 18.7 19.6 18.3 19.5 17.9
Women: 20.0 19.6 21.3 18.5 25.7 30.1 19.6 22.4 20.3 25.9 26.0 15.8
Test at 0.01 level if there is a significant difference in the mean number of minutes used by the
two groups of applicants.
Oh boy, we're testing whether the difference between the mean grades of male and female students is significant. This is gonna be fun!
First, let's state our hypotheses:
H0: There is no significant difference between the means of male and female grades.
H1: There is a significant difference between the means of male and female grades.
To test this, we can use a two-sample t-test. But before we start calculating anything, let me just say that if there's not a significant difference, we might have to resort to a spontaneous dance-off. Just saying.
Alright, let's do some number crunching. The mean grade for males is 79.6, and for females it's 72.125. The standard deviation for males is approximately 8.37, and for females it's around 12.19.
Now, we need to calculate the t-test statistic. Don't worry, I'll do the math. The t-value is approximately 1.05.
Next, we need to find the degrees of freedom. Since we have 10 males and 8 females, the degrees of freedom is 10+8-2=16.
Now, let's consult the t-distribution table. At α=0.01 with 16 degrees of freedom, the critical t-value is approximately 2.921.
Comparing our t-value of 1.05 to the critical t-value of 2.921, we can safely say that the difference between the means of these two samples is not statistically significant. No dance-off required!
So, in conclusion, there's no significant difference between the mean grades of male and female students. But if you still want that dance-off, I'm game!
To test whether the difference between the means of the two samples is significant, we can perform a two-sample t-test.
Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) states that there is no significant difference between the means of the two samples.
The alternative hypothesis (Ha) states that there is a significant difference between the means of the two samples.
Step 2: Determine the significance level:
The significance level (α) is given in the problem as α = 0.01. We will use this value in the calculations.
Step 3: Calculate the sample means and sample standard deviations:
For the male sample:
Mean (X̄m) = (78+87+75+64+90+80+65+86+74+81) / 10 = 78.0
Standard deviation (Sm) = sqrt(((78-78.0)^2 + (87-78.0)^2 + (75-78.0)^2 + ... + (81-78.0)^2) / 9)
= sqrt(168) = 12.964
For the female sample:
Mean (X̄f) = (71+50+80+75+90+61+85+81) / 8 = 73.625
Standard deviation (Sf) = sqrt(((71-73.625)^2 + (50-73.625)^2 + (80-73.625)^2 + ... + (81-73.625)^2) / 7)
= sqrt(196) = 14.000
Step 4: Calculate the test statistic:
The test statistic for a two-sample t-test is given by:
t = (X̄m - X̄f) / sqrt((Sm^2 / n1) + (Sf^2 / n2))
Where n1 = sample size of the male sample, n2 = sample size of the female sample.
In this case, n1 = 10 and n2 = 8.
t = (78.0 - 73.625) / sqrt((12.964^2 / 10) + (14.000^2 / 8))
= 4.375 / 6.885
= 0.635
Step 5: Determine the critical value:
The critical value can be obtained from the t-distribution table using the degrees of freedom, which is (n1 + n2 - 2). In this case, degrees of freedom = 10 + 8 - 2 = 16.
Using a two-tailed test, and α = 0.01, the critical values are ±2.921.
Step 6: Make a decision:
Compare the calculated test statistic (0.635) with the critical values (±2.921).
If the calculated test statistic is outside the critical values, we reject the null hypothesis and conclude that there is a significant difference between the means of the two samples.
If the calculated test statistic is within the critical values, we fail to reject the null hypothesis and cannot conclude that there is a significant difference between the means of the two samples.
In this case, the calculated test statistic (0.635) is within the critical values (-2.921, 2.921). Therefore, we fail to reject the null hypothesis.
Step 7: State the conclusion:
Based on the given data and the test conducted at α = 0.01 level, we do not have sufficient evidence to conclude that there is a significant difference between the means of the two samples.
To test whether the difference between the means of these two samples is significant, we need to perform a hypothesis test.
Here's how you can do it:
Step 1: State the null and alternative hypotheses
- Null Hypothesis (H0): There is no significant difference between the means of the two samples.
- Alternative Hypothesis (H1): There is a significant difference between the means of the two samples.
Step 2: Determine the test statistic
Since we are comparing two independent samples with unknown population variances, we can use the two-sample t-test. The formula for the test statistic is:
t = (mean1 - mean2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where mean1 and mean2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes.
Step 3: Calculate the test statistic
Let's calculate the necessary values to compute the test statistic:
For the male sample:
- Sample mean (mean1) = (78 + 87 + 75 + 64 + 90 + 80 + 65 + 86 + 74 + 81) / 10 = 78.0
- Sample standard deviation (s1) = sqrt([sum of (x - mean1)^2] / (n1 - 1)) = sqrt([(78 - 78)^2 + (87 - 78)^2 + ... + (81 - 78)^2] / 9) = 8.16
For the female sample:
- Sample mean (mean2) = (71 + 50 + 80 + 75 + 90 + 61 + 85 + 81) / 8 = 73.375
- Sample standard deviation (s2) = sqrt([sum of (x - mean2)^2] / (n2 - 1)) = sqrt([(71 - 73.375)^2 + (50 - 73.375)^2 + ... + (81 - 73.375)^2] / 7) = 13.18
Now, let's calculate the test statistic using the formula mentioned above:
t = (78.0 - 73.375) / sqrt((8.16^2 / 10) + (13.18^2 / 8))
= 4.625 / sqrt(6.3804 + 22.3923)
= 4.625 / sqrt(28.7727)
= 4.625 / 5.3685
= 0.8613
Step 4: Determine the critical value and the decision rule
With α = 0.01 and a two-tailed test (because we are looking for a significant difference in either direction), we need to find the critical t-value. Since there are different degrees of freedom for the two samples, we need to use a t-distribution table or a t-distribution calculator.
For a two-tailed test with α = 0.01 and 16 degrees of freedom (10 + 8 - 2), the critical t-value is approximately ±2.921.
Step 5: Make a decision
Compare the calculated test statistic (0.8613) with the critical t-value (±2.921) obtained in step 4. If the calculated test statistic falls within the rejection region (i.e., if it is greater than the positive critical t-value or less than the negative critical t-value), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since the calculated test statistic (0.8613) is within the range of -2.921 to 2.921, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant difference between the means of the two samples at the 0.01 significance level.
In summary, based on the provided data and test results, we cannot conclude that there is a significant difference in the means of the statistics examination grades between male and female students in the very large class.