Question
22102.35 19230.71 10524.63 15574.56 22955.35 20156.87 14146.1 13934.86 18814.42 24729.4 18099.44 18094.57 19855.25 26227.1 10172.86 17326.22 20742.33 18245.59 26793.35 18746.47 15024.9 7850.794 21971.78 30918.98 13486.58
22102.35 |
19230.71 |
10524.63 |
15574.56 |
22955.35 |
20156.87 |
14146.1 |
13934.86 |
18814.42 |
24729.4 |
18099.44 |
18094.57 |
19855.25 |
26227.1 |
10172.86 |
17326.22 |
20742.33 |
18245.59 |
26793.35 |
18746.47 |
15024.9 |
7850.794 |
21971.78 |
30918.98 |
13486.58 |
16443.13 |
17816.94 |
20501.38 |
13403.94 |
24902.91 |
18534.76 |
13200.08 |
9534.121 |
21173.16 |
24454.8 |
19733.54 |
17896.17 |
17298.67 |
17586.39 |
12622.33 |
15519.62 |
29892.31 |
24534.66 |
30868.32 |
22189.32 |
24831.07 |
21282.5 |
19620.42 |
13109.75 |
19029.89 |
17173.21 |
14230.1 |
15564.2 |
20138.13 |
24522.2 |
16347.76 |
17479.89 |
19352.99 |
23319.28 |
23795.24 |
A charity claims that a proportion of 0.75 of donations go to communities where the average family income is less than $20,000 a year. A watchdog group checks a random sample of 60 communities that received funds from the charity to determine if the proportion is different than 0.75 at a significance level of 0.05.
Use Sheet 8 of the Excel file, which gives the average family incomes for the 60 communities, to calculate p^, the z-score, and the p-value
p^=
z=
p=
Since the p-value is Pick greater OR less than the significance level 0.05, the null hypothesis Pick fails to be rejected OR is rejected.
Pick Sufficient OR Insufficient evidence exists that the proportion of donations to communities with an average family income of less than $20,000 a year is different from 0.75.
Step by Step Solution
There are 3 Steps involved in it
Step: 1
Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started