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.