Question 977493: I have most of this, sorry I don't know how to type some of the symbols, i'm stuck at the end need step by step. Thank you!
Confidence interval and hypothesis test
Q-A botanist has produce a new variety of hybrid wheat that is better able to withstand drought than other varieties. She is concerned about whether the seeds will actually grow, so she plants 400 seeds from, the hybrid plant and found that 335 germinate.
A-What is the best point estimate for the proportion of new hybrid seeds that will germinate?
335/400 = 0.8375 or 83.75%
B-Construct a 95% confidence interval to estimate the proportion of hybrid seeds that will germinate?
1-.95=.05/2=.025 Critical Z score =1.959 or 1.96
̂ p hat= .8375 (don’t know how to make the symbols
q hat= 1-.8375 = .1625
n= 400
E=.0361530687
.8375-.0361530682= .8013469318
.8375+.0361530682=.8736530687
Between 80.136% and 87.356% seeds will germinate
C-The botanist knows that for the parent plants, the probability of seeds germination is 80%. The probability of seed germination for the new hybrid variety is unknown, but the botanist claims it is better than 80%. Test the claim that the new hybrid seed has better germination that the old seed. Use a 5% level of significance. State your null hypotheses, test statistic, critical value and the p-value. State the conclusion in laymen terms.
H0:p=.80
H1:p >.80
.05 critical value is 1.645
Test statistic = ?
P value= ?
83.75% is higher than 80.% so i would reject the H0
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You were doing so well until you got to the test statistic.
Your hypotheses are stated correctly:
And your C.V. is stated correctly:
because for ">" in  we are doing a right-tailed test.
The next thing you need to do is check to see if a z-test is appropriate for your proportion test. You need to calculate  and  and both of them must be greater than 5. Well, 20% of 400 is a good deal more than 5, so you are safe.
Your test statistic is then
With the numbers:
For the  -value you need to either use your stat calculator or Excel NORMSDIST function and subtract from 1, or interpolate the value for z = 1.875 and subtract from 1 to get the probability that a z-value is greater than 1.875, in other words the area to the right of 1.875. Excel gives me 0.0304 to four places.
Now we get to the conclusion. You can compare the test statistic to the critical value: 1.875 > 1.645 which tells you, for a right-tailed test, that it is significantly unlikely that the difference in the sample proportion and the population proportion is a result of chance. On this basis alone, you should reject  .
But you also have a -value which is smaller than  , so you can reject on this basis as well. But more than that, instead of only being able to claim that you are 95% sure of your conclusion, you can claim that you are 96.96% sure.
All very well and good for us statisticians, but what about the poor liberal arts schmuck who still has trouble with long division. Let's tell the story in ways that he can understand it:
We have determined that with an approximately 3% margin of error that there is sufficient evidence to reject the notion that the germination proportion of the new variety of hybrid wheat is the same as the 80% proportion experienced with normal varieties of wheat, and can conclude that the new variety has a higher germanation proportion.
John
My calculator said it, I believe it, that settles it
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