Answer :
Based on the evidence from the given sample, we have sufficient evidence to refute the claim that the mean is 40 with a standard deviation of 15.
To determine whether the evidence refutes the claim that the mean is 40 with a standard deviation of 15, we can perform a hypothesis test using the given sample mean and the claimed mean.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The mean is equal to 40 (µ = 40).
Alternative hypothesis (HA): The mean is not equal to 40 (µ ≠ 40).
We can use a t-test to compare the sample mean to the claimed mean. Given that we have a sample size of 100, we can assume the sample mean follows an approximately normal distribution due to the Central Limit Theorem.
The test statistic for this scenario is the t-score, which measures the difference between the sample mean and the claimed mean in terms of standard errors.
The formula for the t-score is:
t = (sample mean - claimed mean) / (sample standard deviation / sqrt(sample size))
Using the given information, we can calculate the t-score:
t = (35.8 - 40) / (15 / sqrt(100))
t = -4.2 / (15 / 10)
t = -4.2 / 1.5
t = -2.8
Next, we need to compare the t-score with the critical value at a given significance level (e.g., α = 0.05) and the degrees of freedom (df = n - 1 = 100 - 1 = 99).
Using a t-table or statistical software, we can find the critical values for a two-tailed test with a significance level of 0.05 and 99 degrees of freedom. The critical t-values are approximately -1.984 and 1.984.
Since the calculated t-score (-2.8) is outside the range of the critical values (-1.984 to 1.984), we reject the null hypothesis.
Therefore, based on the evidence from the given sample, we have sufficient evidence to refute the claim that the mean is 40 with a standard deviation of 15.
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