We want to find a 95 confidence interval for the true

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We want to find a 95% confidence interval for the true proportion ofthe population who favor the bill.What is the lower limit of the interval? (Round to 3 decimal digits)
Hide FeedbackFind the SE (Standard Error). Multiply this by the Critical value of Z to find the ME (Margin ofError). The 68-95-99.7% rule tells us that the critical Z value is 2 (or if you used the Z table, yougot 1.96).Now subtract the ME from the sample proportion to find the lower limit of the populationconfidence interval6 / 6 pointsCompute a 95% confidence interval for the population mean, basedon the sample numbers 21, 28, 33, 34, 25, 26, and 135.Which table do we use to find the margin of error?
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Because the population standard deviation is not known, we use the t-table6 / 6 pointsCompute a 95% confidence interval for the population mean, basedon the sample numbers 21, 28, 33, 34, 25, 26, and 135.What is the Critical value? (Get 4 decimal digits).Use the table here:Distribution TablesQuestion options:
Hide FeedbackT with 6 degrees of freedom. It is a 2-tail test, so we split the .05 rejection region to two and use 0.0256 / 6 pointsCompute a 95% confidence interval for the population mean, basedon the sample numbers 21, 28, 33, 34, 25, 26, and 135.What is the margin of error? (Round to two decimal digits)
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7 / 7 pointsCompute a 95% confidence interval for the population mean, basedon the sample numbers 21, 28, 33, 34, 25, 26, and 135.Find the lower and upper limits of the interval.Question options:Lower - 15.449; Upper - 67.174Lower - 20.492; Upper - 40.876Lower - 18.925; Upper - 50.741Lower - 5.449; Upper - 80.837
Hide FeedbackAdd and subtract the margin of error to/from the mean7 / 7 pointsChange the last value to 27 and re-compute the confidence interval.The numbers now are: 21, 28, 33, 34, 25, 26, and 27
.73; Upper - 51.994What is an outlier and how does it affect the confidence interval?

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