Many dairy cows now receive injections of BST, a hormone intended to spur greater milk production. A group of 53 Jersey cows increased average milk production from 43 to 52 pounds per day, with a standard deviation of 4.8 lbs.
Is this evidence that the hormone maybe effective in this breed of cattle? Assume the assumptions/conditions have been met.
The 95% confidence interval is (, ). Show one decimal places in your answer
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. What level of significance are you using? P = .05?
95% = mean ± 1.96 SEm
paired test hard
To determine if the hormone is effective in increasing milk production in Jersey cows, we can conduct a hypothesis test and calculate a confidence interval.
First, let's state the null and alternative hypotheses:
Null Hypothesis (H₀): The hormone has no effect on milk production in Jersey cows.
Alternative Hypothesis (H₁): The hormone increases milk production in Jersey cows.
Next, we can calculate the test statistic using the formula:
t = (x̄ - μ₀) / (s / √n)
Where:
x̄ = sample mean (average milk production)
μ₀ = hypothesized mean (no effect, in this case)
s = sample standard deviation
n = sample size
Given:
Sample mean (x̄) = 52 lbs/day
Hypothesized mean (μ₀) = 43 lbs/day
Sample standard deviation (s) = 4.8 lbs/day
Sample size (n) = 53 cows
Calculating the test statistic:
t = (52 - 43) / (4.8 / √53)
t ≈ 35.56
With 53 degrees of freedom, the critical value for a 95% confidence interval (two-tailed test) is approximately ±2.009.
The 95% confidence interval is calculated as:
x̄ ± (t * (s / √n))
Using the values we have:
52 ± (2.009 * (4.8 / √53))
52 ± (2.009 * 0.658)
52 ± 1.323
The 95% confidence interval is (50.677, 53.323).
To interpret the results, if the hypothesized mean (43 lbs/day) falls outside the confidence interval (50.677 to 53.323), it would provide evidence that the hormone is effective in increasing milk production in Jersey cows.
To determine whether the hormone BST is effective in increasing milk production in Jersey cows, we can conduct a hypothesis test and calculate a confidence interval.
Step 1: Set up the hypotheses
- Null Hypothesis (H0): The hormone BST does not have any effect on milk production in Jersey cows.
- Alternative Hypothesis (H1): The hormone BST is effective in increasing milk production in Jersey cows.
Step 2: Choose the significance level
The significance level, denoted as α, is the threshold that we set to determine whether there is enough evidence to reject the null hypothesis. Let's assume a significance level of 0.05 (or 5%).
Step 3: Perform the test
To perform the test, we can use a one-sample t-test since we only have data from one group of cows before and after the hormone injection.
The test statistic for this scenario is the t-score. Using the formula for a one-sample t-test:
t = (x̄ - μ) / (s / sqrt(n))
Where:
- x̄: sample mean (average milk production after hormone injection) = 52 lbs
- μ: population mean (average milk production before hormone injection) = 43 lbs
- s: sample standard deviation = 4.8 lbs
- n: sample size = 53 cows
Plugging in the values, we can calculate the t-score.
t = (52 - 43) / (4.8 / sqrt(53))
t ≈ 27.92
Next, we need to find the critical t-value at a 95% confidence level. Since we have a large sample size (n > 30), we can use the standard normal distribution to find the critical t-value. At a 95% confidence level, the critical t-value (two-tailed) is approximately 1.96.
Step 4: Make the decision
Compare the t-score to the critical t-value. If the t-score falls outside the range of -1.96 to 1.96, we can reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.
Since the t-score of 27.92 falls outside the range of -1.96 to 1.96, we can reject the null hypothesis. This provides evidence that the hormone BST is effective in increasing milk production in Jersey cows.
Step 5: Calculate the confidence interval
To calculate the confidence interval, we use the formula:
CI = x̄ ± (t * (s / sqrt(n)))
Using the values we have:
CI = 52 ± (1.96 * (4.8 / sqrt(53)))
Calculating this, the 95% confidence interval is approximately (50.3, 53.7) when rounded to one decimal place.
Therefore, we can say with 95% confidence that the true average increase in milk production for Jersey cows due to the hormone BST lies between 50.3 lbs and 53.7 lbs per day.