Mid Term Question
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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.
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.
Sample mean (`\bar x`): 31 days
Population standard deviation (σ): 4 days
Sample size (n): 40 fields
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.
A researcher claims that the average cost of water bottle is less than 85. He selects a random sample of 36 water bottles from different shops and finds the following costs. Is there enough evidence to support the researcher’s claim at α 0.10? Assume Ï = 22.
60 70 75 55 80 55
50 40 80 70 50 95
120 90 75 85 80 60
110 65 80 85 82 45
75 60 90 90 60 95
110 85 45 90 70 70
Certainly, to test whether there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85, we can conduct a one-sample t-test since the population standard deviation is not provided.
Sample size (n): 36 water bottles
Sample mea n (`\bar x`) and Sample standard deviation (s) are not given directly but can be calculated from the data.
Claimed average cost (μ): $85
Level of significance (α): 0.10
Population standard deviation (σ): $22
First, let's find the sample mean and sample standard deviation from the provided data:
60, 70, 75, 55, 80, 55, 50, 40, 80, 70, 50, 95, 120, 90, 75, 85, 80, 60, 110, 65, 80, 85, 82, 45, 75, 60, 90, 90, 60, 95, 110, 85, 45, 90, 70, 70
Calculating the sample mea n (`\bar x`):
`\bar x = \frac{\sum _{i = 1}^n x_i}{n}`
`\bar x = \frac{3010}{36}`
`\bar x \approx 83.611`
Now, the sample standard deviation (s):
The formula for the sample standard deviation is:
`s = \sqrt {\frac{ \sum _{i = 1}^n (x_i - \bar x )^2}{n - 1}}`
After calculations, the sample standard deviation s ≈ 19.724.
Now, to perform the t-test:
The null and alternative hypotheses are:
Null Hypothesis (H0):μ≥85 (The mean cost is greater than or equal to $85)
Alternate Hypothesis (H1):μ<85 (The mean cost is less than $85)
The test statistic is calculated using the formula:
`t = \frac{\bar x - \mu }{\frac{s}{\sqrt n }}`
`t = \frac{83.611 - 85}{\frac{19.72}{\sqrt 36 }}`
`t = \frac{ - 1.39}{\frac{19.72}{6}}`
t≈−2.111
Degrees of freedom (df) = 1=36−1=35
Using a t-table or statistical software, at a 0.10 level of significance (for a one-tailed test), the critical value for a t-distribution with 35 degrees of freedom is approximately -1.297.
The calculated t-value (-2.111) is less than the critical value (-1.297).
Therefore, we reject the null hypothesis.
This means that there is enough evidence to support the researcher's claim that the average cost of water bottles is less than $85 at a 10% level of significance.
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