Answer :
The absolute value of the test statistic (0.0202) is less than the critical value (2.763), we do not reject the null hypothesis.
Based on the sample data, at a significance level of 0.01, there is not enough evidence to conclude that the sample weights come from a population with a mean different from 150 pounds.
Here, we have,
To test the claim that the sample weights come from a population with a mean equal to 150 pounds, we can perform a one-sample t-test using the traditional method of hypothesis testing.
Given:
Sample size (n) = 11
Sample mean (x) = 149.9 pounds (rounded to one decimal place)
Population mean (μ) = 150 pounds
Population standard deviation (σ) = 16.4 pounds
Hypotheses:
Null Hypothesis (H0): The population mean weight is equal to 150 pounds. (μ = 150)
Alternative Hypothesis (H1): The population mean weight is not equal to 150 pounds. (μ ≠ 150)
Test Statistic:
The test statistic for a one-sample t-test is calculated as:
t = (x - μ) / (σ / √n)
Calculation:
Plugging in the values:
t = (149.9 - 150) / (16.4 / √11)
t ≈ -0.1 / (16.4 / 3.317)
t ≈ -0.1 / 4.952
t ≈ -0.0202
Critical Value:
To determine the critical value at a 0.01 significance level, we need to find the t-value with (n-1) degrees of freedom.
In this case, (n-1) = (11-1) = 10.
Using a t-table or calculator, the critical value for a two-tailed test at a significance level of 0.01 with 10 degrees of freedom is approximately ±2.763.
we have,
Since the absolute value of the test statistic (0.0202) is less than the critical value (2.763), we do not reject the null hypothesis.
we get,
Based on the sample data, at a significance level of 0.01, there is not enough evidence to conclude that the sample weights come from a population with a mean different from 150 pounds.
Learn more about standard deviation here:
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