Answer:
A.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. Z=2.255. P=0.01207.
C. Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. P(z>0.94)=0.1736
Step-by-step explanation:
A. state the null and alternative hypotheses.
The null hypothesis states that the review course has no effect, so the scores are still the same. The alternative hypothesis states that the review course increase the score.
[tex]H_0: \mu\leq514\\\\H_1: \mu>514[/tex]
B. test the hypothesis at the a=.10 level of confidence. is a mean math score of 520 significantly higher than 514? find the test statistic, find P-value. is the sample statistic significantly higher?
The test statistic Z can be calculated as
[tex]Z=\frac{M-\mu}{s/\sqrt{N}} =\frac{520-514}{119/\sqrt{2000}}=\frac{6}{2.661}=2.255[/tex]
The P-value of z=2.255 is P=0.01207.
The P-value is smaller than the significance level, so the effect is significant. The null hypothesis is rejected.
Then we can conclude that the score of 520 is significantly higher than 514, in this case, specially because the big sample size.
C. do you think that a mean math score of 520 vs 514 will affect the decision of a school admissions adminstrator? in other words does the increase in the score have any practical significance?
Although it is not big difference, it is an improvement that has evidence. The scores are expected to be higher in average than without the review course.
D. test the hypothesis at the a=.10 level of confidence with n=350 students. assume that the sample mean is still 520 and the sample standard deviation is still 119. is a sample mean of 520 significantly more than 514? find test statistic, find p value, is the sample mean statisically significantly higher? what do you conclude about the impact of large samples on the p-value?
In this case, the z-value is
[tex]Z=\frac{520-514}{s/\sqrt{n}} =\frac{6}{119/\sqrt{350}} =\frac{6}{6.36} =0.94\\\\P(z>0.94)=0.1736>\alpha[/tex]
In this case, the P-value is greater than the significance level, so there is no evidence that the review course is increasing the scores.
The sample size gives robustness to the results of the sample. A large sample is more representative of the population than a smaller sample. A little difference, as 520 from 514, but in a big sample leads to a more strong evidence of a change in the mean score.
Large samples give more extreme values of z, so P-values that are smaller and therefore tend to be smaller than the significance level.
A math teacher claims to have developed a review course that increases the score of students on the math portion of a college entrance exam. The null hypothesis is that there is no difference in scores before and after the course, while the alternative hypothesis is that the course increases the scores. The hypothesis is tested at the 0.10 level of confidence by calculating the z-score and finding the p-value. The impact of the score increase on the decision of a school admissions administrator depends on their criteria. The hypothesis is then tested again with a larger sample size, following the same steps as earlier.
Explanation:A. The null hypothesis (H0) is that there is no difference in scores before and after the review course, while the alternative hypothesis (H1) is that the review course increases the scores on the math portion of the college entrance exam.
B. To test the hypothesis at the 0.10 level of confidence, we calculate the z-score and find the p-value. The test statistic is z = (sample mean - population mean) / (population standard deviation / sqrt(sample size)). The p-value is the probability of obtaining a test statistic as extreme as the observed value, assuming the null hypothesis is true. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the mean math score is significantly higher than 514.
C. Whether a mean math score increase from 514 to 520 has any practical significance depends on the requirements set by the school admissions administrator. If the administrator determines that a small increase in score does not significantly impact their decision, then the increase may not have practical significance. However, if the administrator considers even small improvements in score as valuable, then the increase may have practical significance.
D. To test the hypothesis at the 0.10 level of confidence with a sample size of 350 students, we follow the same steps as in part B to calculate the test statistic and p-value. If the p-value is less than the significance level, we reject the null hypothesis and conclude that the sample mean of 520 is significantly higher than 514. Large sample sizes can result in smaller p-values, indicating stronger evidence against the null hypothesis.
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