Problem #2. Information on a packet of seeds claims that 94% of them will germin

Question

Problem #2. Information on a packet of seeds claims that 94% of them will germinate. In order to take a look at the reality, we took a sample of 200 seeds and checked the status of germination as follows: where 1 means the seed is germinated and 0 means the seed is not germinated. Based on this data, please answer the following questions. 1. what is the three components of a 95% confidence interval for the true proportion of seeds that germinate based on this sample? 2. Put the three components in a confidence interval form and interpret the result. 3. Does this seem to provide evidence that the claim is wrong? 4. Repeat the parts (1)- (3) under 90% confidence level. 5. Instead of decreasing confidence level, what would be another approach you can at- tempt for to show the claim is wrong?

Explanation / Answer

1)

n =200 , z value at 95% = 1.96 , p = 180/200 = 0.9

2)

n = 200

p = 0.9

z-value of 95% CI = 1.9600

SE = sqrt(p*(1-p)/n) = 0.02121

ME = z*SE = 0.04158

Lower Limit = p - ME = 0.85842

Upper Limit = p + ME = 0.94158

95% CI (0.8584 , 0.9416 )

we are 95% confident that the true proportion of seeds that germinate based on this sample is between 0.8584 and 0.9416

3)

As 94% is covered in the CI, it seems that claim is not wrong.

4.

n = 200

p = 0.9

z-value of 90% CI = 1.6449

SE = sqrt(p*(1-p)/n) = 0.02121

ME = z*SE = 0.03489

Lower Limit = p - ME = 0.86511

Upper Limit = p + ME = 0.93489

90% CI (0.8651 , 0.9349)

we are 90% confident that the true proportion of seeds that germinate based on this sample is between 0.8651 and 0.9349

As 94% is not covered in the CI, it seems that claim is wrong.

5.

Another approach is to increase the sample size.

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