(a) A report included the following information on the heights (in.) for non-His

Question

(a)

A report included the following information on the heights (in.) for non-Hispanic white females.

Calculate a confidence interval at confidence level approximately 95% for the difference between population mean height for the younger women and that for the older women. (Use ?20–39 ? ?60 and older.)

( _____ , _____ )

Let ?1 denote the population mean height for those aged 20–39 and ?2 denote the population mean height for those aged 60 and older. Interpret the hypotheses H0: ?1 ? ?2 = 1 and Ha: ?1 ? ?2 > 1. The null hypothesis states that the true mean height for younger women is 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women. Carry out a test of these hypotheses at significance level 0.001. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

  z=

P-value=

(b)

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm2/min) was measured. For m = 10 subjects with the syndrome, the average heat output was x = 0.62, and for n = 10 nonsufferers, the average output was 2.07. Let ?1 and ?2 denote the true average heat outputs for the sufferers and nonsufferers, respectively. Assume that the two distributions of heat output are normal with ?1 = 0.3 and ?2 = 0.4. Consider testing H0: ?1 ? ?2 = ?1.0 versus Ha: ?1 ? ?2 < ?1.0 at level 0.01. Describe in words what Ha says, and then carry out the test. Ha says that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers.

  z =

p-value =

State the conclusion in the problem context. Reject H0. The data suggests that the average heat output for sufferers is more than 1 cal/cm2/min below that of non-sufferers. What is the probability of a type II error when the actual difference between ?1 and ?2 is ?1 ? ?2 = ?1.3? (Round your answer to four decimal places.)

___________

Assuming that m = n, what sample sizes are required to ensure that ? = 0.1 when ?1 ? ?2 = ?1.3? (Round your answer up to the nearest whole number.)

___________ subjects

(c)

Tensile strength tests were carried out on two different grades of wire rod, resulting in the accompanying data.

Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2? Test the appropriate hypotheses using a significance level of 0.01. State the relevant hypotheses. H0: ?1064 ? ?1078 = ?10 Ha: ?1064 ? ?1078 < ?10. Calculate the test statistic and P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Reject H0. The data suggests that the true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm2. Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability. (Use a 95% confidence interval. Round your answers to two decimal places.)

( ______ , ______ ) kg/mm^2

(d)

An experiment was performed to compare the fracture toughness of high-purity 18 Ni maraging steel with commercial-purity steel of the same type. For m = 30 specimens, the sample average toughness was x = 65.1 for the high-purity steel, whereas for n = 37 specimens of commercial steel y = 59.6. Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial-purity steel by more than 5. Suppose that both toughness distributions are normal. Assuming that ?1 = 1.4 and ?2 = 1.1, test the relevant hypotheses using ? = 0.001. (Use ?1 ? ?2, where ?1 is the average toughness for high-purity steel and ?2 is the average toughness for commercial steel.) H0: ?1 ? ?2 = 5 Ha: ?1 ? ?2 > 5. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Fail to reject H0. The data does not suggest that the fracture toughness of high-purity steel exceeds that of commercial-purity steel by more than 5. Compute ? for the test conducted in part (a) when ?1 ? ?2 = 6. (Round your answer to four decimal places.)

? =

Age Sample
Size Sample
Mean Std. Error
Mean 20–39 867 64.7 0.09 60 and older 934 63.0 0.11

Explanation / Answer

There are way too amny questions posted in here in a single question. I shall try ot altest answer 4 sub-querries.

a). The hypothesis model will be

Ho:?1 - ?2 = 1

Ha:?1 - ?2 > 1

The test statistic is given by

t = [ (x1 - x2) - d ] / SE (x1,x2 are sample means, d =difference between them =5,SE =standard error given by

SE = sqrt[ (s12/n1) + (s22/n2) ] where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Keeping values we have

SE =0.1421

t=(64.7-63.0-1)/0.1421

=4.9261

for t=4.9261 DF=1792

The P-Value is < .00001.

The result is significant at p < .05.

b). The hypothesis model will be

Ho:?1 - ?2 = 1

Ha:?1 - ?2 < 1

The test statistic is given by

t = [ (x1 - x2) - d ] / SE (x1,x2 are sample means, d =difference between them =5,SE =standard error given by

SE = sqrt[ (s12/n1) + (s22/n2) ] where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Keeping values we have

SE =0.158

t=(2.07-0.62-1)/0.1580

=2.8481

for t=2.8481 DF=17

The P-Value is .005559.

The result is significant at p < .01.

Now when

Ho:?1 - ?2 = 1.3

Ha:?1 - ?2 < 1.3

then t=(2.07-0.62-1.3)/0.1580

=0.9493

The P-Value is .177885.

The result is not significant at p < .01.

The probalbity of type !! erro will be 17.78 %

c). The hypothesis model will be

Ho:?1 - ?2 = 10

Ha:?1 - ?2 < 10

The test statistic is given by

t = [ (x1 - x2) - d ] / SE (x1,x2 are sample means, d =difference between them =5,SE =standard error given by

SE = sqrt[ (s12/n1) + (s22/n2) ] where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Keeping values we have

SE =0.2174

t=(124.5-101.6-10)/0.2174

=59.333

for t=59.333 DF=187

The P-Value is < .00001.

The result is significant at p < .01.

d). The hypothesis model will be

Ho:?1 - ?2 = 5

Ha:?1 - ?2 > 5

The test statistic is given by

t = [ (x1 - x2) - d ] / SE (x1,x2 are sample means, d =difference between them =5,SE =standard error given by

SE = sqrt[ (s12/n1) + (s22/n2) ] where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Keeping values we have

SE =0.313

t=(65.1-59.6-5)/0.313

=1.5968

for t=1.5968   DF=54

The P-Value is .058073.

The result is not significant at p < .01.

Now When

The hypothesis model will be

Ho:?1 - ?2 = 6

Ha:?1 - ?2 > 6

The t statisitc will be

t=(65.1-59.6-5)/0.313

=0.5968

The P-Value is .276357.

The result is not significant at p < .01.

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