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2 months ago
 Need help with the formula 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?b.   Use the five steps of hypothesis testing. c.   Figure the confidence limits for the 95% confidence interval. Read 55 times 2 Replies
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wrote...
2 months ago
 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? Use the five steps of hypothesis testing.Figure the confidence limits for the 95% confidence interval.Quote1. Was the average test score earned by the first 20 students to turn in their tests significantly different from the overall mean?NO2. (73.62;82.38) for the 95% confidence interval.
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2 months ago
 The provided sample mean is ​X​¯​​ = 78 and the known population standard deviation is σ=10, and the sample size is n = 20(1) Null and Alternative HypothesesThe following null and alternative hypotheses need to be tested:Ho: μ = 75Ha: μ≠75This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.(2) Rejection RegionBased on the information provided, the significance level is α=.05, and the critical value for a two-tailed test is z​c​​= 1.96The rejection region for this two-tailed test is R={z:∣z∣>1.96}(3) Test StatisticsThe z-statistic is computed as follows:z=​σ/√​n​​​​​​X​¯​​−μ​0​​​​=​10/√20​​78−75​​ = 1.342(4) Decision about the null hypothesisSince it is observed that ∣z∣=1.342≤z​c​​=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) ConclusionIt 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 IntervalThe 95% confidence interval is given asCI=​x​¯​​±z​α​​∗s/√n=78±1.96∗10/√20=73.617<μ<82.383.Explanation:for any further clarification let me know
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