Question 1110153
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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