Assume x1 = 104, x2 = 128, s1 = 11, s2-12, n1 = 50, and n2-50. Find a 95% confid

Question

Assume x1 = 104, x2 = 128, s1 = 11, s2-12, n1 = 50, and n2-50. Find a 95% confidence interval for the difference in the corresponding values off. (Round your answers to three decimal places.) Does this interval include more or fewer values than a 99% confidence interval? Explain your answer This interval includes fewer values because a smaller confidence level means that we need a smaller margin of error O This interval includes fewer values because a smaller confidence level means that we need a larger margin of error. This interval includes more values because a smaller confidence level means that we need a smaller margin of error O This interval includes more values because a smaller confidence level means that we need a larger margin of error

Explanation / Answer

we know that CI si given as

statistic + margin of error.

now the z value is 1.96 for the 95% CI , from the z table

When the standard deviation of either population is unknown and the sample sizes (n1 and n2) are large, the standard deviation of the sampling distribution can be estimated by the standard error, using the equation below.

SEx1-x2 = sqrt [ s21 / n1 + s22 / n2 ]

putting the values sqrt(121/50 + 144/50) = 2.30

now difference in means is x1bar-x2bar = 104-128 = -24

the confidence interval is thus

-24 +- 1.96*2.30

-19.5 , -28.5

the inerval would include more values because we need a smaller margin of error , hence C

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