A researcher wishes to check the impact of fertilizer on yield | Query Point Official
A researcher wishes to check the impact of fertilizer on a yield, estimated time period to grow a yield in a field is 28 days. A sample of 40 fields have been taken and a mean of 31 days. At α = 0.05, test the claim that the mean time is greater than 28 days. The standard deviation of the population is 4 days.
Solution:
To test the claim that the mean time to grow a yield with the use of fertilizer is greater than 28 days, you can perform a one-sample z-test.
Given:
Sample mean (`\bar x`): 31 days
Population standard deviation (σ): 4 days
Sample size (n): 40 fields
Hypotheses:
Null Hypothesis ( H0): μ≤28 (Mean time is less than or equal to 28 days)
Alternate Hypothesis (H1): μ>28 (Mean time is greater than 28 days)
Level of significance (α): 0.05
The z-score formula for the sample mean is:
`Z = \frac{\bar x - \mu }{\frac{\sigma }{\sqrt n }}`
Where:
`\bar x` = Sample mean
μ = Population mean
σ = Population standard deviation
n = Sample size
First, calculate the z-score:
`Z = \frac{31 - 28}{\frac{4}{\sqrt {40} }}`
`Z = \frac{3}{\frac{4}{\sqrt {40} }}`
`Z = \frac{3 \times \sqrt {40} }{4}`
Z≈7.66
Next, we will find the critical value for a one-tailed test with a significance level of 0.05.
At a significance level of 0.05, using a standard normal distribution table or a z-table, the critical z-value is approximately 1.645.
As the calculated z-value (7.66) is greater than the critical value (1.645), we reject the null hypothesis.
Therefore, based on the given sample data, there is sufficient evidence to conclude that the mean time to grow a yield with the use of fertilizer is significantly greater than 28 days at a 5% level of significance.
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Frequently Asked Questions (FAQs)
What type of hypothesis test is used here?
A one-sample z-test is used because the population standard deviation is known and the sample size is large.
What does the significance level α mean?
The significance level α = 0.05 means we accept a 5% chance of incorrectly rejecting the null hypothesis if it is true.
Why is this a one-tailed test?
Because the alternative hypothesis specifically tests if the mean is greater than a value (28 days), not just different.
What conclusion is drawn from rejecting the null hypothesis?
Rejecting H0 indicates sufficient evidence that fertilizer increases the crop growth time above 28 days.
When should a t-test be used instead of a z-test?
A t-test is used when the population standard deviation is unknown and must be estimated from the sample.
Related Topics
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