Question

suppose we dont know the population standard deviation σ. to test the null hypothesis ( h_0: mu = 10 ), we take a random sample of size 25 and calculate the sample mean ( overline{x} = 11 ) and the sample standard deviation ( s = 2 ). what is the p - value for this test? hint: you can use the online p - value calculator .012419 .019343 .019654 .621629

Explanation:

Step1: Calculate the t - statistic

The formula for the t - statistic in a one - sample t - test is \(t=\frac{\bar{X}-\mu}{s/\sqrt{n}}\).
Given \(\bar{X} = 11\), \(\mu = 10\), \(s = 2\), and \(n = 25\).
First, calculate \(s/\sqrt{n}=\frac{2}{\sqrt{25}}=\frac{2}{5} = 0.4\).
Then, \(t=\frac{11 - 10}{0.4}=\frac{1}{0.4}=2.5\).

Step2: Determine the degrees of freedom

The degrees of freedom for a one - sample t - test is \(df=n - 1\).
Since \(n = 25\), \(df=25-1 = 24\).

Step3: Find the p - value

Using a t - distribution table or an online p - value calculator (for a one - tailed test, assuming the alternative hypothesis \(H_1:\mu>10\)), with \(t = 2.5\) and \(df = 24\).
Looking up the value in the t - distribution: the p - value is the probability that \(T>2.5\) where \(T\sim t(24)\).
Using an online calculator or software (e.g., in R: 1 - pt(2.5,24)), we find the p - value.

Answer:

.0107 (using more precise calculation, but among the given options, if we assume a two - tailed test was intended (though the problem statement is a bit ambiguous, but re - calculating for two - tailed: \(P(|T|>2.5)\) with \(df = 24\), using 2*(1 - pt(2.5,24)) in R gives approximately \(0.019654\)). So the answer is.019654.