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MATH 225N Week 7 Assignment; Conduct a Hypothesis Test for Proportion; P-Value Approach

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1. Question: What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.)…. z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.0380.0460.0570.0690.090.0290.0370.0460.0560.068 2. Question: Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session, Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance level, does the data provide sufficient evidence to conclude that Mary's mean throw is less than 61 meters? Accept or reject the hypothesis given the sample data below. • H0:μ=61 meters; Ha:μ61 meters • α=0.01 (significance level) • z0=−1.99 • p=0.0233 • The Right Choice Is Option E • i.e • Do Not Reject The Null Hypothesis because The P value is 0.02333 is Greater than The Significance Level 0.01 3. Question: What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27? (Do not round your answer; compute your answer using a value from the table below.)….. z0.10.20.30.40.50.000.5400.5790.6180.6550.6910.010.5440.5830.6220.6590.6950.020.5480.5870.6260.6630.6980.030.5520.5910.6290.6660.7020.040.5560.5950.6330.6700.7050.050.5600.5990.6370.6740.7090.060.5640.6030.6410.6770.7120.070.5670.6060.6440.6810.7160.080.5710.6100.6480.6840.7190.090.5750.6140.6520.6880.722 4. Question: Marty, a typist, claims that his average typing speed is 72 words per minute. During a practice session, Marty has a sample typing speed mean of 84 words per minute based on 12 trials. At the 5% significance level, does the data provide sufficient evidence to conclude that his mean typing speed is greater than 72 words per minute? Accept or reject the hypothesis given the sample data below. • H0:μ≤72 words per minute; Ha:μ72 words per minute • α=0.05 (significance level) • z0=2.1 • p=0.018 5. Question: Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session, Nancy has a sample driving distance mean of 229.6 yards based on 18 drives. At the 2% significance level, does the data provide sufficient evidence to conclude that Nancy's mean driving distance is less than 253 yards? Accept or reject the hypothesis given the sample data below. • H0:μ=253 yards; Ha:μ253 yards • α=0.02 (significance level) • z0=−0.75 • p=0.2266 6. Question: Kathryn, a golfer, has a sample driving distance mean of 187.3 yards from 13 drives. Kathryn still claims that her average driving distance is 207 yards, and the low average can be attributed to chance. At the 1% significance level, does the data provide sufficient evidence to conclude that Kathryn's mean driving distance is less than 207 yards? Given the sample data below, accept or reject the hypothesis. • H0:μ=207 yards; Ha:μ207 yards • α=0.01 (significance level) • z0=−1.46 • p=0.0721 7. Question: What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.01? (Do not round your answer; compute your answer using a value from the table below.)….. z−1.2−1.1−1.0−0.90.000.1150.1360.1590.1840.010.1130.1330.1560.1810.020.1110.1310.1540.1790.030.1090.1290.1520.1760.040.1070.1270.1490.1740.050.1060.1250.1470.1710.060.1040.1230.1450.1690.070.1020.1210.1420.1660.080.1000.1190.1400.1640.090.0990.1170.1380.161 8. Question: What is the p-value of a left-tailed one-mean hypothesis test, with a test statistic of z0=−1.19?.... Use a p-value rounded to 3 decimal places…… Here is a portion of the Standard Normal Table. To read the Standard Normal table, match the ones and tenths digits of the z-value in the first column with the correct hundredths digit in the first row. 9. Question: What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.59? (Do not round your answer; compute your answer using a value from the table below.)…. z−1.8−1.7−1.6−1.5−1.40.000.0360.0450.0550.0670.0810.010.0350.0440.0540.0660.0790.020.0340.0430.0530.0640.0780.030.0340.0420.0520.0630.0760.040.0330.0410.0510.0620.0750.050.0320.0400.0490.0610.0740.060.0310.0390.0480.0590.0720.070.0310.0380.0470.0580.0710.080.0300.0380.0460.0570.0690.090.0290.0370.0460.0560.068 10. Question: Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and record their annual salary. Gina calculates the sample mean income to be $56,500 per year with a sample standard deviation of 3,750. Using the alternative hypothesis Ha:μ55,000, find the test statistic t and the p-value for the appropriate hypothesis test. Round the test statistic to two decimal places and the p-value to three decimal places.

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Conduct a Hypothesis Test for Proportion - P-
Value Approach
Week 7
1. A teacher claims that the proportion of students expected to pass an
exam is greater than 80%. To test this claim, the teacher administers the
test to 200 random students and determines that 151 students pass the
exam.

The following is the setup for this hypothesis test:

H0:p=0.80
Ha:p>0.80
In this example, the p-value was determined to be 0.944.
Come to a conclusion and interpret the results for this hypothesis test for a
proportion (use a significance level of 5%)
The decision is to fail to reject the Null Hypothesis.
The conclusion is that there is not enough evidence to support the claim.

2. A hospital administrator claims that the proportion of knee surgeries that
are successful is 87%. To test this claim, a random sample of 450 patients
who underwent knee surgery is taken and it is determined that 371 of these
patients had a successful knee surgery operation.

The following is the setup for this hypothesis test:

H0:p=0.87
Ha:p≠0.87
In this example, the p-value was determined to be 0.004.
Come to a conclusion and interpret the results for this hypothesis test for a
proportion (use a significance level of 5%)
the decision is to reject the Null Hypothesis.
The conclusion is that there is enough evidence to reject the claim.

, 3. A college administrator claims that the proportion of students that are
nursing majors is greater than 40%. To test this claim, a group of 400
students are randomly selected and its determined that 190 are nursing
majors.

The following is the setup for this hypothesis test:

H0:p=0.40

Ha:p>0.40

Find the p-value for this hypothesis test for a proportion and round your
answer to 3 decimal places.

The following table can be utilized which provides areas under the Standard
Normal Curve:

0.001.

4. A police officer claims that the proportion of accidents that occur in the
daytime (versus nighttime) at a certain intersection is 35%. To test this
claim, a random sample of 500 accidents at this intersection was examined
from police records it is determined that 156 accidents occurred in the
daytime.

The following is the setup for this hypothesis test:

H0:p = 0.35

Ha:p ≠ 0.35

Find the p-value for this hypothesis test for a proportion and round your
answer to 3 decimal places.

0.076
5. A teacher claims that the proportion of students expected to pass an
exam is greater than 80%. To test this claim, the teacher administers the

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