The world’s smallest mammal is the bumblebee bat. Such bats are roughly
the same size of a large bumblebee. Listed below are the weights (in
grams) from a sample of these bats. Assuming that all such bats have a
standard deviation of .3 grams, use a significance level of .05 to test the
claim that these bats are from the same population with a known mean of
1.8 grams. Do the bats appear to be from the same population ? You may
assume the sample data comes from a population that follows a normal
distribution.
2.13 1.31 1.91 1.92 1.77 1.64 2.29 1.99 2.33 2.49 2.22
Find the mean of your sample first = sum of scores/number of scores
Subtract each of the scores from the mean and square each difference. Find the sum of these squares. Divide that by the number of scores to get variance.
Standard deviation = square root of variance
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 your Z score.
To test the claim that the bats are from the same population with a known mean of 1.8 grams, we can use a one-sample t-test. Here's how we can perform this test:
Step 1: State the null and alternative hypotheses:
The null hypothesis (H0) assumes that the mean weight of the bats is equal to 1.8 grams.
The alternative hypothesis (Ha) assumes that the mean weight of the bats is not equal to 1.8 grams.
H0: μ = 1.8 (where μ is the population mean weight)
Ha: μ ≠ 1.8
Step 2: Choose the level of significance:
The question specifies a significance level of 0.05, which means we are willing to accept a 5% chance of making a Type I error (rejecting the null hypothesis when it is true).
Step 3: Calculate the test statistic:
To calculate the test statistic, we can use the formula for a one-sample t-test:
t = (x̄ - μ) / (s / √n)
Where:
x̄ is the sample mean,
μ is the known population mean,
s is the sample standard deviation,
n is the sample size.
For this problem, we know that μ = 1.8 grams, and the sample data is as follows:
2.13, 1.31, 1.91, 1.92, 1.77, 1.64, 2.29, 1.99, 2.33, 2.49, 2.22
Calculating the sample mean (x̄):
x̄ = (2.13 + 1.31 + 1.91 + 1.92 + 1.77 + 1.64 + 2.29 + 1.99 + 2.33 + 2.49 + 2.22) / 11
x̄ ≈ 1.988 grams
Calculating the sample standard deviation (s):
s = √[(Σ(x - x̄)^2) / (n - 1)]
s ≈ 0.355 grams
Calculating the test statistic (t):
t = (1.988 - 1.8) / (0.355 / √11)
t ≈ 2.238
Step 4: Determine the critical value and p-value:
Since we have a two-tailed test (μ ≠ 1.8), we need to consider both tails of the t-distribution.
Using a significance level of 0.05 and degrees of freedom (df) = n - 1 = 10, we can find the critical values using a t-table or statistical software. In this case, the critical values are approximately -2.228 and 2.228.
We can also use statistical software or a t-distribution calculator to find the p-value associated with the test statistic. In this case, the p-value is approximately 0.046.
Step 5: Make a decision:
If the test statistic falls within the critical regions (outside the critical values) or if the p-value is less than the significance level (0.05), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, the test statistic (t ≈ 2.238) falls outside the critical values (-2.228, 2.228), and the p-value (0.046) is less than the significance level (0.05). Therefore, we reject the null hypothesis.
Conclusion:
Based on the provided data and using a significance level of 0.05, there is evidence to suggest that the bats do not appear to be from the same population with a known mean weight of 1.8 grams.