Question

A.

x 8.4.6-T Question Help A simple random sample of pulse rates of 25 women from a normally distributed population results in a standard deviation of 10.6 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.01 significance level to test the claim that pulse rates of women have a standard deviation equal to 10 beats per minute. Complete parts (a) through (d) below. a. Identify the null and alternative hypotheses. Choose the correct answer below. O A. Ho: 0= 10 beats per minute Hy: 0 < 10 beats per minute c. Ho: a = 10 beats per minute Hy: 0# 10 beats per minute O B. Ho: 6 2 10 beats per minute H4:0< 10 beats per minute OD. Ho: 0# 10 beats per minute H: 0 = 10 beats per minute b. Compute the test statistic. x=0 (Round to three decimal places as needed.) Enter your answer in the answer box and then click Check Answer. parts 2 remaining Clear All Check Answer

Compute the test statistic x²⁼_____ (Round to three decimal places)

Find the P-value=______ (Round to three decimal places)

Reject/Fail To reject _____ H0. There is/is not_____ sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.

B.

Find the test statistic= x^{2=_____ (Round to three decimal places)}

Find the p-value of test statistic=_______ (^{Round to three decimal places)}

(1) _________H0. There(2)__________sufficient evidence to conclude that the tar content of filtered​ 100-mm cigarettes has a standard deviation different from 0.25 mg.

(1)Fail to reject/Reject

(2)is/is not

C.

Calculate the value of the test statistic.

X2 = _____ (Round to three decimal places as needed.)

Use technology to determine the? P-value for the test statistic.

The? P-value is _______ (Round to three decimal places as needed.)

What is the correct conclusion at the ?=0.10 level of? significance?

Since the? P-value is ____ than the level of? significance, _______ the null hypothesis. There _____ sufficient evidence to conclude that the standard deviation of heights of? major-league baseball players is less for?major-league baseball players at the 0.10 level of significance.

D.

Find the test statistic=_______​(Round to two decimal places as​ needed.)
Determine the critical​ value(s).

Determine the critical value

The critical​ value(s) is/are= ________ ​(Use a comma to separate answers as needed. Round to two decimal places as​ needed.)

Since the test statistic is _____ the critical​ value(s), ______ Upper H 0. There is ______ evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft. The variation appears to be _____ than in the​ past, so the new method appears to be _______​, because there will be _______ altimeters that have errors.​ Therefore, the company _______ take immediate action to reduce the variation.
▼
less
greater than
about the same
▼

Reject/Fail to reject

sufficient/not sufficient

similar
better
worse
▼
the same number of
fewer
more
▼
should
should not

P.S PLEASE PLEASE HELP ME SOLVE PART A-D. MY ASSIGNMENTS DUE TONIGHT AND I NEED THESE TO BE COMPLETED TO BE ABLE TO PASS CLASS. IM BEGGING YOU PLEASE HELP.

Answer 1

A.

s = 10.6, n = 25

α = 0.01

Null and alternative hypothesis:

Hₒ : σ = 10

H₁ : σ ≠ 10

Test statistic:

χ² = (n-1)s² / σ² = (25 - 1)10.6² /10² = 26.966

Degree of freedom:

Df = n -1 = 25 - 1 = 24

p-value:  

p-value = CHISQ.DIST.RT(26.966, 24) = 0.3060

Decision:

p-value > α, fail to reject the null hypothesis.

There is not sufficient evidence to warrant rejection of the claim that pulse rates of women have a standard deviation equal to 10 beats per minute.

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B.

s = 0.18, n = 28

α = 0.05

Null and alternative hypothesis:

Hₒ : σ = 0.25

H₁ : σ ≠ 0.25

Test statistic:

χ² = (n-1)s² / σ² = (28 - 1)0.18² /0.25² = 13.997

Degree of freedom:

Df = n -1 = 28 - 1 = 27

p-value = 1-CHISQ.DIST.RT(13.9968, 27) = 0.0187

Reject H0. There is sufficient evidence to conclude that the tar content of filtered​ 100-mm cigarettes has a standard deviation different from 0.25 mg.

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C.

Sample standard deviation using excel function STDEV.S, s = 2.0800

Sample size, n = 20

Null and alternative hypothesis:

Hₒ : σ = 2.6

H₁ : σ < 2.6

Test statistic:

χ² = (n-1)s² / σ² = (20 - 1)2.08² /2.6² = 12.160

Degree of freedom:

Df = n -1 = 20 - 1 = 19  

p-value = 1-CHISQ.DIST.RT(12.1598, 19) = 0.1213

Since the P-value is more than the level of significance, Fail to reject the null hypothesis. There is not sufficient evidence to conclude that the standard deviation of heights of major-league baseball players is less for major-league baseball players at the 0.10 level of significance.

-----------------

D.

Sample standard deviation using excel function STDEV.S, s = 58.3928

n = 12

Test statistic:

χ² = (n-1)s² / σ² = (12 - 1)58.3928² /32.2² = 36.174

Degree of freedom:

Df = n -1 = 12 - 1 = 11

Critical value, χ²α = CHISQ.INV.RT(0.05, 11) = 19.68

Since the test statistic is greater than the critical​ value(s), reject Upper H0. There is sufficient evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft. The variation appears to be worse than in the​ past, so the new method appears to be worse​, because there will be more altimeters that have errors.​ Therefore, the company should not take immediate action to reduce the variation.