Question 1110153: A researcher interested in whether people are able to identify emotions correctly in other people when they are extremely tired. It is known that, using a particular method of measurement, the accuracy rating of people in the general population (who are not extremely tired) are normally distributed with a mean of 82 and a variance of 20. In the present study, however, the researcher arranges to test 50 people who had no sleep the previous night. The mean accuracy for these 50 individuals was 78. Using a two-tailed test and the .05 significance level, what should the researcher conclude? (a) Carry out a Z test using the five steps of hypothesis testing. (b)Make a drawing of the distribution involved. (c) Explain your answer to someone who knows about hypothesis testing with a sample of a single individual but who knows nothing about hypothesis testing with a sample of more than one individual. (d) ADVANCED TOPIC: Figure the 95% confidence interval and explain your answer to someone who is familiar with the general logic of hypothesis testing, the normal curve, Z scores, probability, and the idea of a distribution of means, but who has not heard of confidence interval
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The researcher postulates there is no difference between the population mean and the sample mean at the 0.05 significance level
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Ho: sample mean = 82
H1: sample mean not = 82
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Since the sample size is > 30, we can us the normal distribution
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the test statistic is (78 - 82)/(20/square root(50)) = −1.4142
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the probability associated with this test statistic is 0.0793
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since this is a two tailed test we calculate 0.0793 + 0.0793 = 0.1586 as the p-value
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since the p-value is > 0.05, the researcher accepts the null hypothesis which is sleep deprivation does not affect the person's ability to discern emotions in another person
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a) In a two-tailed test the critical value for 0.05 level of significance is -1.960 and 1.960
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Note in a two-tailed test we divide the significance level by 2 and use the result(in this case 0.025) as the probability to look up the critical z-value
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we know the test statistic is -1.4142, therefore we accept the null hypothesis(Ho) since -1.4141 is > -1.960
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b) it's a normal curve, along the x-axis show 0 in the middle and 1, 2, 3 to the right of 0 and -1, -2, -3 to the left. on the x-axis show -1.96 and 1.96 then draw two vertical lines from the -1.960 to intersect the normal curve and from the 1.960 to intersect the normal curve. label both of these areas a/2 = 0.025
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c) we assume the general population is normally distributed and since our sample is > 30 individuals we can use a normal distribution, if the sample size is < 30, we would use the student's t-statistic
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d) we want to construct a 95% confidence interval
the standard error(SE) = 20/square root(50) = 2.8284
alpha(a) = 1 - (95/100) = 0.05
critical probability(p*) = 1 - (a/2) = 0.975
express the critical value(CV) as a z-score, find the z-score having a cumulative probability equal to p* = 0.975
CV = 1.960
the margin of error = CV * SE = 1.960 * 2.8284 = 5.5437
95% confidence interval is 78 + or - 5.5437, interval notation is (72.4563, 83.5437)
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Suppose we used the same sampling method to select different samples and to compute a different interval estimate for each sample. Some interval estimates would include the true population mean(82) and some would not. A 95% confidence level means that we would expect 95% of the interval estimates to include the population mean(82).
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