Problem
A genetic experiment with peas resulted in one sample of offspring that consisted of 435 green peas and 162 yellow peas.
a. Construct a $95 \%$ confidence interval to estimate of the percentage of yellow peas.
b. Based on the confidence interval, do the results of the experiment appear to contradict the expectation that $25 \%$ of the offspring peas would be yellow?
a. Construct a $95 \%$ confidence interval. Express the percentages in decimal form.
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Solution
Step 1 ：Given that there are 162 yellow peas and a total of 597 peas, we can calculate the sample proportion (p) as the number of yellow peas divided by the total number of peas, which is approximately 0.271.
Step 2 ：We use a z-score of 1.96 for a 95% confidence level.
Step 3 ：We calculate the standard error (se) using the formula \(se = \sqrt{\frac{p(1-p)}{n}}\), which gives us approximately 0.018.
Step 4 ：We can then construct the 95% confidence interval using the formula \(p \pm z \times se\). This gives us a lower limit of approximately 0.236 and an upper limit of approximately 0.307.
Step 5 ：Final Answer: The $95 \%$ confidence interval for the proportion of yellow peas is \(\boxed{0.236, 0.307}\).
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Solution
Step 1 ：Given that there are 162 yellow peas and a total of 597 peas, we can calculate the sample proportion (p) as the number of yellow peas divided by the total number of peas, which is approximately 0.271.
Step 2 ：We use a z-score of 1.96 for a 95% confidence level.
Step 3 ：We calculate the standard error (se) using the formula \(se = \sqrt{\frac{p(1-p)}{n}}\), which gives us approximately 0.018.
Step 4 ：We can then construct the 95% confidence interval using the formula \(p \pm z \times se\). This gives us a lower limit of approximately 0.236 and an upper limit of approximately 0.307.
Step 5 ：Final Answer: The $95 \%$ confidence interval for the proportion of yellow peas is \(\boxed{0.236, 0.307}\).
