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2-3AssigmentH.docxRunning Head: ASSIGMENT 2-3: HYPOTHESIS TESTING 1
ASSIGMENT 2-3: HYPOTHESIS TESTING
A. Example 1
Solution in no.1 Problem in minitab finding
|
Significance Level (x) |
55(given) |
|
Standard Deviation (O) |
286.2(determined using Minitab ) |
|
Mean Time (x) |
.025 (given) |
|
Sample Size (x) |
30 (given) |
|
Reject Null Hypothesis if X> P- Valve |
|
|
Null hypothesis μ |
275 |
|
Alternative hypothesis u |
>275 |
In this problem Jeffery’s father assumed that his son will be faster by the use of new expensive google. The null hypothesis is rejected, as you can see on the solution the X (.05) is > than the P value of .019. It means that Jeffery can swim faster than 16.43 seconds with the use of google that receive form his father.
B. Example 2
Solution: Problem in Minitab finding
|
Significance Level (x) |
55(given) |
|
Standard Deviation (O) |
286.2(determined using Minitab ) |
|
Mean Time (x) |
.025 (Given) |
|
Sample Size (x) |
30 (given) |
|
Reject Null Hypothesis if X> P- Valve < .025 |
|
|
Null hypothesis μ |
275 |
In conclusion in no. 2 there is no enough evidence to prove that the players have more than 275 pounds. It says here that the null cannot be rejected in this case P is .132 > X Valve of .025.
C. Example 3
Solution: Problem in Minitab finding
Unknown deviation
In this problem the statistic data says the 05 being > P valve .04 with rejected in null hypothesis. The average score is >65 as per teacher’s prediction.
D. Example 4
By the use of normal approximation, P value is .549.
There is no evidence that the brides are younger to their grooms that are diff. than 50%. The 99% level of confidence can’t capable to reject by the null hypothesis in the value X valve= 0.1 < the P value is .549.
E. Example 5
They have no evidence to tell of the percentage of the household have 3 phones is not 30 percent. The X<P is the answer to the null hypothesis can’t reject.
Reference
Majaski, Christina. (Feb 2019). Hypothesis Testing. Retrieved from
https://www.investopedia.com/terms/h/hypothesistesting.asp
null hypothesis U =65
alternative hypothesis U > 65
Rejected null hypothesis if X > P - Valve
Test of μ 65 vs > 65
variableNMeanSt DevSE Mean 95%L BoundTP
C110673.21.0165.151.980.04
significan level (x).01 (given)
sample size (n)100 (given)
alternative hypothesis p ≠ 50
null hypothesis of p0.5
test and C1 for 1 Proportiontest of P0.5 vs ≠ .05
SampleXNSample P99%CIZ valuePvalue
1531000.530.4014441, 06585590.60.549
Significance Level (n):.05(not given, inferred)
Sample Size (n): 150 (given)
null hypothesis of p0.3
alternative hypothesis p ≠ .30
test of p = 0.3 vs p ≠ 0.3
samplexnsample p95% CIz valuep value
1431500.2866670.214300 And 0.359033-0.360.722
test of μ =16.43 vs 16.43Assumed deviation0.8
NMeanSE Mean 95%Upper BoundZP
15160.20716.34-2.080.019
test of μ 275 vs>275Standard assumed deviation55
NMean SE MEAN 97.5%Lower BoundzP
30286.210266.51.12 0.1230.123
Segnificance of level (x).05 (given)
Mean Score67 (determined by Minitab)
Sample size (n)10 (Given)
No Standard given deviation
