Refer to the data in the accompanying table for the heights of females. Complete parts (a) through (d) below. E Click the icon to view the table of height data. a. Enter the observed frequencies in the table below. Height (cm)Less than 155.65 155.65 - 161.75 Frequency 161.75 - 167.85 Greater than 167.85 (Type whole numbers.) b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, find the probability of a randomly selected height belonging to each class. Height (cm) Less than 155.65 155.65 - 161.75 161.75 Probability 167.85 Greater than 167.85 (Round to four decimal places as needed.) c. Using the probabilities found in part (b), find the expected frequency for each category. Height (cm) Expected Frequency Less than 155.65 155.65 – 161.75 161.75 - 167.85 Greater than 167.85 (Round to two decimal places as needed.) d. Use a 0.05 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population? Identify the null and alternative hypotheses for this test.
Refer to the data in the accompanying table for the heights of females. Complete parts (a) through (d) below. E Click the icon to view the table of height data. a. Enter the observed frequencies in the table below. Height (cm)Less than 155.65 155.65 - 161.75 Frequency 161.75 - 167.85 Greater than 167.85 (Type whole numbers.) b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, find the probability of a randomly selected height belonging to each class. Height (cm) Less than 155.65 155.65 - 161.75 161.75 Probability 167.85 Greater than 167.85 (Round to four decimal places as needed.) c. Using the probabilities found in part (b), find the expected frequency for each category. Height (cm) Expected Frequency Less than 155.65 155.65 – 161.75 161.75 - 167.85 Greater than 167.85 (Round to two decimal places as needed.) d. Use a 0.05 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population? Identify the null and alternative hypotheses for this test.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
GENDER (1 equals M) | HEIGHT |
0 | 171.6 |
1 | 184.4 |
0 | 153.2 |
1 | 161.8 |
1 | 179.6 |
1 | 165.9 |
1 | 178.5 |
1 | 157.2 |
0 | 156.1 |
1 | 181.4 |
1 | 166.2 |
0 | 167.7 |
1 | 179.9 |
1 | 171.1 |
0 | 163.1 |
0 | 158.2 |
1 | 180.8 |
0 | 155.1 |
0 | 179.9 |
1 | 173.4 |
1 | 172.1 |
1 | 176.1 |
1 | 155.6 |
1 | 181.4 |
0 | 150.4 |
1 | 183.8 |
1 | 170.5 |
0 | 155.9 |
0 | 157.2 |
1 | 161.8 |
0 | 155.8 |
0 | 156.9 |
0 | 155.8 |
0 | 154.9 |
0 | 162.8 |
1 | 165.9 |
0 | 159.8 |
0 | 163.5 |
0 | 165.7 |
0 | 169.2 |
1 | 154.6 |
1 | 179.7 |
1 | 175.2 |
0 | 154.3 |
0 | 160.3 |
0 | 167.9 |
1 | 175.1 |
0 | 157.1 |
0 | 170.3 |
1 | 179.4 |
1 | 183.2 |
0 | 170.6 |
0 | 152.2 |
0 | 160.5 |
0 | 157.2 |
1 | 171.3 |
1 | 174.1 |
0 | 172.4 |
0 | 161.5 |
0 | 164.5 |
0 | 171.2 |
0 | 153.9 |
1 | 183.2 |
1 | 180.5 |
0 | 135.7 |
1 | 187.7 |
1 | 175.5 |
1 | 167.8 |
0 | 167.7 |
1 | 178.8 |
1 | 173.6 |
0 | 160.1 |
0 | 156.4 |
0 | 167.9 |
1 | 188.2 |
0 | 157.1 |
0 | 176.8 |
1 | 179.1 |
1 | 178.7 |
1 | 180.7 |
1 | 176.3 |
0 | 153.1 |
1 | 180.1 |
0 | 163.3 |
0 | 160.5 |
1 | 171.1 |
0 | 167.5 |
1 | 171.3 |
0 | 159.9 |
0 | 155.6 |
0 | 158.2 |
1 | 166.6 |
1 | 167.6 |
1 | 174.5 |
0 | 168.9 |
0 | 156.2 |
1 | 178.7 |
0 | 171.5 |
1 | 170.8 |
1 | 172.7 |
1 | 173.5 |
1 | 184.6 |
1 | 169.8 |
0 | 160.6 |
1 | 167.6 |
0 | 157.9 |
1 | 165.8 |
0 | 161.6 |
1 | 176.9 |
1 | 168.2 |
0 | 154.7 |
1 | 163.4 |
1 | 164.8 |
1 | 171.4 |
1 | 168.9 |
0 | 165.4 |
1 | 157.4 |
1 | 173.2 |
0 | 149.1 |
0 | 152.5 |
0 | 176.1 |
0 | 165.3 |
0 | 154.7 |
0 | 161.8 |
1 | 183.2 |
0 | 165.7 |
0 | 167.1 |
0 | 165.7 |
0 | 167.5 |
0 | 172.8 |
0 | 159.8 |
1 | 180.4 |
0 | 164.5 |
0 | 161.1 |
0 | 162.5 |
0 | 163.2 |
1 | 167.9 |
1 | 182.4 |
0 | 151.5 |
1 | 179.8 |
1 | 167.7 |
0 | 170.1 |
1 | 178.1 |
1 | 165.8 |
1 | 178.5 |
1 | 167.8 |
0 | 160.1 |
1 | 162.4 |
0 | 172.5 |
0 | 161.8 |
1 | 174.6 |
1 | 168.8 |
0 | 159.6 |
0 | 181.8 |
1 | 172.4 |
0 | 157.7 |
0 | 165.3 |
0 | 169.2 |
1 | 161.1 |
1 | 185.3 |
1 | 167.3 |
1 | 159.6 |
1 | 172.2 |
1 | 170.6 |
1 | 164.6 |
0 | 153.3 |
0 | 167.6 |
1 | 166.5 |
0 | 155.4 |
1 | 174.4 |
0 | 156.2 |
1 | 170.1 |
1 | 169.4 |
0 | 161.6 |
0 | 170.5 |
0 | 156.5 |
0 | 162.7 |
0 | 156.1 |
0 | 162.7 |
1 | 183.4 |
1 | 166.1 |
1 | 169.6 |
1 | 166.3 |
0 | 156.6 |
1 | 168.8 |
0 | 160.2 |
0 | 169.1 |
1 | 180.1 |
1 | 176.5 |
1 | 182.3 |
1 | 172.5 |
0 | 167.1 |
0 | 157.7 |
1 | 178.5 |
1 | 168.8 |
1 | 168.1 |
1 | 187.3 |
1 | 171.3 |
1 | 180.2 |
0 | 155.9 |
1 | 169.6 |
0 | 156.9 |
0 | 167.5 |
1 | 169.5 |
1 | 167.8 |
1 | 186.2 |
0 | 145.1 |
1 | 176.9 |
1 | 170.2 |
0 | 151.1 |
0 | 158.7 |
0 | 159.3 |
1 | 175.8 |
0 | 162.9 |
1 | 178.6 |
0 | 149.3 |
0 | 165.6 |
1 | 170.4 |
1 | 172.9 |
0 | 165.4 |
0 | 169.7 |
1 | 148.6 |
0 | 161.6 |
1 | 165.3 |
1 | 164.2 |
0 | 164.6 |
0 | 163.7 |
0 | 162.3 |
1 | 162.8 |
![Refer to the data in the accompanying table for the heights of females. Complete parts (a) through (d) below.
E Click the icon to view the table of height data.
a. Enter the observed frequencies in the table below.
Height (cm) Less than 155.65 155.65 – 161.75
161.75 – 167.85 Greater than 167.85
Frequency
(Type whole numbers.)
b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, find the probability of a randomly selected height belonging to each class.
Height (cm) Less than 155.65 155.65 – 161.75
161.75 – 167.85
Greater than 167.85
Probability
(Round to four decimal places as needed.)
c. Using the probabilities found in part (b), find the expected frequency for each category.
Height (cm) Less than 155.65 155.65 – 161.75
Expected
Frequency
161.75 – 167.85
Greater than 167.85
(Round to two decimal places as needed.)
d. Use a 0.05 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population?
Identify the null and alternative hypotheses for this test.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d1239aa-d77b-4fd4-9256-17d7501a8b6d%2F15070f5b-3ec1-4dfe-b2c9-118eb444e1bf%2Fgixvpfd_processed.png&w=3840&q=75)
Transcribed Image Text:Refer to the data in the accompanying table for the heights of females. Complete parts (a) through (d) below.
E Click the icon to view the table of height data.
a. Enter the observed frequencies in the table below.
Height (cm) Less than 155.65 155.65 – 161.75
161.75 – 167.85 Greater than 167.85
Frequency
(Type whole numbers.)
b. Assuming a normal distribution with mean and standard deviation given by the sample mean and standard deviation, find the probability of a randomly selected height belonging to each class.
Height (cm) Less than 155.65 155.65 – 161.75
161.75 – 167.85
Greater than 167.85
Probability
(Round to four decimal places as needed.)
c. Using the probabilities found in part (b), find the expected frequency for each category.
Height (cm) Less than 155.65 155.65 – 161.75
Expected
Frequency
161.75 – 167.85
Greater than 167.85
(Round to two decimal places as needed.)
d. Use a 0.05 significance level to test the claim that the heights were randomly selected from a normally distributed population. Does the goodness-of-fit test suggest that the data are from a normally distributed population?
Identify the null and alternative hypotheses for this test.
![Identify the test statistic.
(Round to three decimal places as needed.)
Identify the P-value
(Round to three decimal places as needed.)
State the final conclusion that addresses the original claim.
Ho. There is
V evidence to warrant rejection of the claim that the female heights were randomly selected from a normally distributed population.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d1239aa-d77b-4fd4-9256-17d7501a8b6d%2F15070f5b-3ec1-4dfe-b2c9-118eb444e1bf%2Fawit3vi_processed.png&w=3840&q=75)
Transcribed Image Text:Identify the test statistic.
(Round to three decimal places as needed.)
Identify the P-value
(Round to three decimal places as needed.)
State the final conclusion that addresses the original claim.
Ho. There is
V evidence to warrant rejection of the claim that the female heights were randomly selected from a normally distributed population.
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