Eleven students were randomly selected from a population of 1000 students. The sampling method was simple random sampling. All of the students were given a standardized English test and a standardized math test. Test results are summarized below. Student English Math 1 95 90 2 89 85 3 76 73 4 92 90 5 91 90 6 53 53 7 67 68 8 88 90 9 75 78 10 85 89 11 90 95 (a) Find the 90% confidence interval for the mean difference between student scores on the math and English tests. Assume that the mean differences are approximately normally distributed. Round confidence interval to nearest thousandth. (0.12765 would be entered as 0.128) Lower , Upper (b) Is there any difference between the mean scores on the math and English tests at α=0.10 ?
Eleven students were randomly selected from a population of 1000 students. The sampling method was simple random sampling. All of the students were given a standardized English test and a standardized math test. Test results are summarized below.
| Student | English | Math | ||
| 1 | 95 | 90 | ||
| 2 | 89 | 85 | ||
| 3 | 76 | 73 | ||
| 4 | 92 | 90 | ||
| 5 | 91 | 90 | ||
| 6 | 53 | 53 | ||
| 7 | 67 | 68 | ||
| 8 | 88 | 90 | ||
| 9 | 75 | 78 | ||
| 10 | 85 | 89 | ||
| 11 | 90 | 95 |
(a) Find the 90% confidence interval for the
Round confidence interval to nearest thousandth. (0.12765 would be entered as 0.128)
Lower , Upper
(b) Is there any difference between the mean scores on the math and English tests at α=0.10 ?
Let X represents English and Y represents Math
So mean of X, X bar = summation Xi /n
= (95+89+76+92+91+53+67+88+75+85+90)/11
= 81.91
Variance of X, (Sx )2= summation (Xi2)-(mean)2
= (95^2+89^2+76^2+92^2+91^2+53^2+67^2+88^2+75^2+85^2+90^2)/11 - (81.91)^2
= 6859.91-6709.2481
=150.6619
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