A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Texas and Indiana. Due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana. To see if this is true, the firm selected a random sample of 20 childcare workers from Texas and an independent random sample of 20 childcare workers from Indiana and asked them to report their mean annual income. The data obtained were as follows. Annual income in dollars Texas 37093 , 33960 , 35692 , 29528 , 30978 , 35504 , 26961 , 28380 , 37537 , 25341 , 34837 , 36489 , 33633 , 41229 , 43577 , 27644 , 43364 , 28415 , 28919 , 34096 Indiana 33672 , 46452 , 35900 , 28486 , 46528 , 34252 , 40952 , 42164 , 36298 , 38828 , 32690 , 29157 , 42770 , 35527 , 40804 , 51823 , 37543 , 25530 , 37239 , 43782 Send data to calculator Send data to Excel The population standard deviation for the annual incomes of childcare workers in Texas and in Indiana are estimated as 6200 and 6500 , respectively. It is also known that both populations are approximately normally distributed. At the 0.10 level of significance, is there sufficient evidence to support the claim that the mean annual income, μ1 , of childcare workers in Texas is less than the mean annual income, μ2 , of childcare workers in Indiana? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternative hypothesis H1 . H0: H1: (b) Determine the type of test statistic to use. ▼(Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the critical value at the 0.10 level of significance. (Round to three or more decimal places.) (e) Can we support the claim that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana? Yes No

A nationwide job recruiting firm wants to compare the annual incomes for childcare workers in Texas and Indiana. Due to recent trends in the childcare industry, the firm suspects that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana. To see if this is true, the firm selected a random sample of 20 childcare workers from Texas and an independent random sample of 20 childcare workers from Indiana and asked them to report their mean annual income. The data obtained were as follows. Annual income in dollars Texas 37093 , 33960 , 35692 , 29528 , 30978 , 35504 , 26961 , 28380 , 37537 , 25341 , 34837 , 36489 , 33633 , 41229 , 43577 , 27644 , 43364 , 28415 , 28919 , 34096 Indiana 33672 , 46452 , 35900 , 28486 , 46528 , 34252 , 40952 , 42164 , 36298 , 38828 , 32690 , 29157 , 42770 , 35527 , 40804 , 51823 , 37543 , 25530 , 37239 , 43782 Send data to calculator Send data to Excel The population standard deviation for the annual incomes of childcare workers in Texas and in Indiana are estimated as 6200 and 6500 , respectively. It is also known that both populations are approximately normally distributed. At the 0.10 level of significance, is there sufficient evidence to support the claim that the mean annual income, μ1 , of childcare workers in Texas is less than the mean annual income, μ2 , of childcare workers in Indiana? Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.) (a) State the null hypothesis H0 and the alternative hypothesis H1 . H0: H1: (b) Determine the type of test statistic to use. ▼(Choose one) (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the critical value at the 0.10 level of significance. (Round to three or more decimal places.) (e) Can we support the claim that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana? Yes No

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

childcare workers from Texas and an independent random sample of

childcare workers from Indiana and asked them to report their mean annual income. The data obtained were as follows.

	Annual income in dollars
Texas	37093 , 33960 , 35692 , 29528 , 30978 , 35504 , 26961 , 28380 , 37537 , 25341 , 34837 , 36489 , 33633 , 41229 , 43577 , 27644 , 43364 , 28415 , 28919 , 34096
Indiana	33672 , 46452 , 35900 , 28486 , 46528 , 34252 , 40952 , 42164 , 36298 , 38828 , 32690 , 29157 , 42770 , 35527 , 40804 , 51823 , 37543 , 25530 , 37239 , 43782

Send data to calculator

Send data to Excel

The population standard deviation for the annual incomes of childcare workers in Texas and in Indiana are estimated as

6200

and

6500

, respectively. It is also known that both populations are approximately normally distributed. At the

0.10

level of significance, is there sufficient evidence to support the claim that the mean annual income,

μ1

, of childcare workers in Texas is less than the mean annual income,

μ2

, of childcare workers in Indiana? Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.)

(a)	State the null hypothesis H0 and the alternative hypothesis H1 .
H0:
H1:
(b)	Determine the type of test statistic to use.
	▼(Choose one)
(c)	Find the value of the test statistic. (Round to three or more decimal places.)

(d)	Find the critical value at the 0.10 level of significance. (Round to three or more decimal places.)

(e)	Can we support the claim that the mean annual income of childcare workers in Texas is less than the mean annual income of childcare workers in Indiana?
Yes No

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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