Researchers studying the learning of speech often compare measurements made on t

Question

Researchers studying the learning of speech often compare measurements made on the recorded speech of adults and children. One variable of interest is called the voice onset time (VOT). Here are the results for 6-year-old children and adults asked to pronounce the word "bees." The VOT is measured in milliseconds and can be either positive or negative. Group n x s Children 10 -3.67 33.89 Adults 20 -23.17 50.74

a. What is the standard error of the sample mean VOT for the 20 adult subjects?

b. What is the standard error of the difference xchildren - xadults between the mean VOT for children and adults?

c. The researchers were investigating whether VOT distinguishes adults from children. State H0 and Ha and carry out a two-sample t-test. Give a P-value and report your conclusions.

d. Give a 95% confidence interval for the difference in mean VOTs when pronouncing the word "bees." Explain why you knew from your result in (c) that this interval would contain 0 (no difference). 10 -3.67 33.89 20 -23.17 50.74 Researchers studying the learning of speech often compare measurements made on the recorded speech of adults and children. One variable of interest is called the voice onset time (VOT). Here are the results for 6-year-old children and adults asked to pronounce the word "bees." The VOT is measured in milliseconds and can be either positive or negative. Group n x s Children 10 -3.67 33.89 Adults 20 -23.17 50.74

a. What is the standard error of the sample mean VOT for the 20 adult subjects?

b. What is the standard error of the difference xchildren - xadults between the mean VOT for children and adults?

c. The researchers were investigating whether VOT distinguishes adults from children. State H0 and Ha and carry out a two-sample t-test. Give a P-value and report your conclusions.

d. Give a 95% confidence interval for the difference in mean VOTs when pronouncing the word "bees." Explain why you knew from your result in (c) that this interval would contain 0 (no difference).

Explanation / Answer

a.

std. error of the adult subjects = 50.74/sqrt(20) = 11.3458

b.

SE = sqrt[ (s1^2/n1) + (s2^2/n2) ]

SE = sqrt((33.89^2/10 + 50.74^2/20)

SE = 15.6071

c.

H0: mu1 - mu2 = 0

Ha: mu1 - mu2 not equals to 0

test statistic, t = (-3.67 - -23.17)/15.6071

t = 1.2494

p-value = 0.2430 (df = 9)

As p-value is greater than significance level of 0.05, we fail to reject the null hypothesis.

there is no difference in the VOT

d)

x1bar - x2bar = 19.50

SE = 15.6071

CI = 95%

DF = 9

t-value = 2.2622

ME = t*SE = 35.3056

Confidence Interval is (x1bar - x2bar) +/- ME

Lower bound = 19.50 - 35.3056 = -15.80564

Upper bound = 19.50 + 35.3056 = 54.80564

Confidence Interval is (-15.8056 , 54.8056)

In part c we fail to reject the null hypothesis, which means 0 is lies within the CI.

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