Question

a study timed left-handed and right-handed people to see how long it took them to hit a buzzer. the data is shown below: the sixty right handed people had an average of 2.4 s and a standard deviation of 0.2 s. the fifty nine left handed people had an average of 2.5 s and a standard deviation of 0.5 s. the difference is -0.1 s, and the standard deviation of the matched pairs differences is 0.2 s. find an 87% confidence interval for the difference in left-handed or right-handed people. what z or t score should you use?

Answer 1

The z or t score to be used is  (0.422,0.978)

How to find the critical value ?

Critical probability (p*) = 1 - (Alpha / 2), where Alpha is equal to 1 - (the confidence level / 100), is the unit statisticians use to determine the margin of error within a set of data in statistics.

The critical value for α = 0.13 is $z_{c} = \frac{z_{1} - \alpha }{2}$

The corresponding confidence interval is computed as shown below

$CI = (X1_{1} - X_{2}- z_{c} \sqrt{\frac{a_{1}^{2}}{n_{1} } + \frac{a_{2}^{2}}{n_{2} }} , X1_{1} - X_{2}+z_{c} \sqrt{\frac{a_{1}^{2}}{n_{1} } + \frac{a_{2}^{2}}{n_{2} }} )$

$CI = (2.1 - 1.4- 1.514\sqrt{\frac{1^{2}}{37 } + \frac{0.5_{2}^{2}}{38 }} , 2.1 - 1.4 + 1.514 \sqrt{\frac{1^{2}}{37 } + \frac{0.5^{2}}{38 }} )$

CI = (0.422, 0.978)

0.422< mu1 - mu 2 < 0.978

To learn more about finding the critical value from the given link  

brainly.com/question/14040224

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