Answer :
Since our calculated t-statistic (2.33) is less than the critical t-value (2.36), we fail to reject the null hypothesis (H0).
To test the claim, we can conduct a t-test for the mean. The null and alternative hypotheses are:
H0: μ = 529 (mean total score is 529)
H1: μ ≠ 529 (mean total score is different from 529)
Using the sample data, we calculate the sample mean (x) and sample standard deviation (s):
x = (699 + 560 + 414 + 570 + 521 + 663 + 727 + 413) / 8 = 586.5
s = √[(Σ(x - x)²) / (8 - 1)] = 121.4
Next, we calculate the t-statistic and degrees of freedom:
t = (x - 529) / (s / √8)
= (586.5 - 529) / (121.4 / √8)
= 2.33
df = 8 - 1
= 7
Using the t-distribution table or calculator, we find the critical t-value for α = 0.05 and df = 7:
t(0.05, 7) = ±2.36
Since our calculated t-statistic (2.33) is less than the critical t-value (2.36), we fail to reject the null hypothesis (H0). This means we cannot conclude that the mean total score is significantly different from 529 at the 0.05 level.
The correct option is c. There is not enough evidence to reject the claim that the mean total score is different from 529.
To test the claim that the mean total score is different from 529, we will conduct a one-sample t-test. The null hypothesis (H0) is that the population mean [tex](\(\mu\))[/tex] is equal to 529, and the alternative hypothesis (H1) is that [tex]\(\mu\)[/tex] is not equal to 529.
Given the sample of 8 students' scores: 699, 560, 414, 570, 521, 663, 727, 413.
First, we calculate the sample mean [tex](\(\bar{x}\))[/tex]:
Next, we calculate the sample standard deviation (s):
[tex]\[s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}\]\[[/tex]
[tex]\[s = \sqrt{\frac{128^2 + (-10.875)^2 + (-156.875)^2 + (-0.875)^2 + (-49.875)^2 + (92.125)^2 + (156.125)^2 + (-157.875)^2}{7}}\][/tex]
[tex]\[s = \sqrt{\frac{16489 + 118.30078125 + 24539.0625 + 0.765625 + 2485.0625 + 8482.1875 + 24395.6875 + 24934.80078125}{7}}\][/tex]
[tex]\[s = \sqrt{17616.378348214}\][/tex]
[tex]\[s \approx 419.75\][/tex]
Now, we calculate the t-statistic:
With [tex]\(n-1\)[/tex] degrees of freedom, we look up the critical t-value for a two-tailed test with [tex]\(\alpha = 0.05\)[/tex][tex]\(\alpha = 0.05\)[/tex]. For 7 degrees of freedom, the critical t-value is approximately 2.365.
Since the absolute value of our calculated t-statistic (0.282) is less than the critical t-value (2.365), we fail to reject the null hypothesis.
Therefore, there is not enough evidence to reject the claim that the mean total score is different from 529 at the 5% significance level.
