Researchers studying the acquisition of pronunciation often compare measurements made on the recorded speech of adults and children. One variable of interest is called "voice onset time" (VOT), the length of time between the release of a consonant sound (such as "p") and the beginning of an immediately following vowel (such as the "a" in "pat"). For speakers of English, this short time lag can be heard as a period of breathiness between the consonant and the vowel. Here are the results for some randomly selected 4-year-old children and adults asked to pronounce the word "pat". VOT is measured in milliseconds and can be either positive or negative.

Children: n = 10, mean = 60.67, standard deviation = 39.89

Adults: n = 20, mean = 88.17, standard deviation = 24.74

You are interested in whether there is a difference in the VOT of adults and children, so you plan to test H0:μa−μc=0 against Ha:μa−μc≠0, where μa and μcare the population mean VOT for adults and children, respectively.

A. What additional information would you need to confirm that the conditions for this test have been met?

B. Assuming the conditions have been met, calculate the test statistic and p-value for this test.

C. Interpret the p-value in the context of this study and draw the appropriate conclusion at
α = 0.05.

D. Given your conclusion in part C, which type error, Type I or Type II, is it possible to make? Describe that error in the context of this study.

(A) The additional information that is needed to confirm about the conditions for this test have been met is ‘Population is approximately normal.

(B) The test statistic = 1.9965

(C1) P-value = 0.0556

(C2) There is no significant difference between mean VOT for children and adults.

D1) It is possible to make type II error.

(D2) There should be a difference in the VOT of adults and children.

Step-by-step explanation:

Let na be the number of adults = 20

xa mean VOT for adults = 88.17

sa standard deviatiation of VOT for adults = 24.74

Let nc be the number of children = 10

xc mean VOT for children =60.67

sc standard deviatiation of VOT for children = 39.89

(A) From the information the population variances are unknown and the two sample are assumed to be independent and the sample the sample size are smaller that is (n<30).

This indicates that the additional information that is required for the conditions of the test to be satisfied is 'distribution of the population'. the addition assumption to be made is, that the population distribution is normal.

(b) Calculating the test statistics using the formula;

t = (xa -xc)/SE - d

where SE = standard deviation , d= hypothesized difference = 0

But SE = √sa²/na +sc²/nc

           = √24.74 ²/20 + 39.89²/10

           = √189.72559

           = 13.774

Substituting into test statistics equation, we have

t = (xa -xc)/SE - d

  = (88.17 - 60.67/13.774

  = 1.9965

Therefore the test statistic is 1.9965

(c) Calculating the p-value, we have;

Degree of freedom = na + nc -2

                                 = 10+ 20 -2

                                 = 28

The p-value for t=1.9965 at 28 degrees of freedom and 0.05 level of significance is 0.0556.

The p-value 0.0556 is greater than given level of significance 0.05 hence we fail to reject the null hypothesis and conclude that there is no significant difference between mean VOT for children and adults.

(D1) From the information in part (C) the null hypothesis is not rejected.

Since the null hypothesis is not rejected, there might be a chance that not rejecting null hypothesis would be wrong. In this type of situation the error that can occur would be type II error.

(D2) The type of error describe in the context of this study is obtained by the concept of the type II error which tells that the null hypothesis is not rejected when it is actually false.

7

step-by-step explanation:

the answer would be 7

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.