Josephine is taking a college course, and her professor bases the course grade on the average of 2 tests taken in the semester.
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Answer:
The null hypothesis:
The alternative hypothesis:
Test statistics : z = −2.30
P-value: = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27
Step-by-step explanation:
From the given information:
Let's state the null and the alternative hypothesis;
Since The claim is that 27% of the offspring peas will be yellow.
The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.
i.e
The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27
<u>The test statistics:</u>
we are given 437 green peas and 129 yellow apples;
Hence;
where ;
= sample proportion
x = number of success
n = total number of the sample size
Now; the test statistics can be computed as :
z = −2.30
C. P-value
P-value = P(Z < z)
P-value = P(Z< -2.30)
By using the P-value method and the normal distribution as an approximation to the binomial distribution.
from the table of standard normal distribution
move left until the first column is reached. Note the value as –2.0
move upward until the top row is reached. Note the value as 0.30
find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of
z = -2.30
P- value = 2P(z ≤ -2.30)
P-value = 2 × 0.01072
P - value = 0.02144
Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.
Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27