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Title: 6.A psychology professor of a large class became curious as to whether the students who turned in t Post by: brendalee on Jul 23, 2019 6. A psychology professor of a large class became curious as to whether the students who turned in tests first scored differently from the overall mean on the test. The overall mean score on the test was 75 with a standard deviation of 10; the scores were approximately normally distributed. The mean score for the first 20 students to turn in tests was 78. Using the .05 significance level, was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?
a. Use the five steps of hypothesis testing. Title: Re: 6.A psychology professor of a large class became curious as to whether the students who turned ... Post by: jmendoza on Jul 23, 2019 Hi brendalee (https://biology-forums.com/index.php?action=profile;u=854153)
Null hypothesis: Mean test score for students who turned in the test first = overall test mean Alternative hypothesis: Mean test score for students who turned in the test first ≠ overall test mean Significance level: .05 Test statistic / critical value: t-value of 1.96 (note that this critical value is given for a sample of infinite size. Because n of the overall sample has not been provided, t cannot be precisely calculated) Conclusion: Reject the null hypothesis if the |t| of the comparison between the two means < 1.96 Confidence limits for the 95% confidence interval Assuming n = 100, the 95% confidence interval is given by the formula of 75 ± (2.04)(1.8) = 71.33 to 78.67 [Note that the 95% confidence interval for the entire sample cannot be perfectly estimated because n is not given, and the 95% confidence interval for the students who turned in their test first cannot be calculated because the standard deviation for this sample has not been given]. Title: Re: 6.A psychology professor of a large class became curious as to whether the students who turned ... Post by: samyflashy on Jul 23, 2019 The provided sample mean is 78, and the known population standard deviation is σ=10, and the sample size is n = 20
(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ = 75 Ha: μ≠75 This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is α=.05, and the critical value for a two-tailed test is z = 1.96 The rejection region for this two-tailed test is R={z:∣z∣>1.96} (3) Test Statistics The z-statistic is computed as follows: z = 1.342 (4) Decision about the null hypothesis Since it is observed that ∣z∣=1.342≤z =1.96 , it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.1797, and since p=0.1797≥.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is different than 75, at the .05 significance level. Confidence Interval The 95% confidence interval is given as 73.617<μ<82.383. Title: Re: 6.A psychology professor of a large class became curious as to whether the students who turned ... Post by: bio_man on Jul 23, 2019 In this post, https://biology-forums.com/index.php?topic=139802.0, A and D are answered and accepted as best answer too!
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