Question

A population (with mean μ) follows normal distribution. Ten samples (N) are drawn at random with a mean value of "x" and standard deviation of "S". Following table provides the confidence limits, C(t) of the cumulative probability function for Student's t - distribution two-tailed test with degree of freedom, D.
Which one of the following expression is correct for testing the null hypothesis $H_o: μ = 0$ at 10% significance level?

D	C(t)
	0.9	0.95	0.975
9	1.38	1.83	2.26
10	1.37	1.81	2.23
11	1.36	1.80	2.20

• A. \(-181 < \frac{x}{\frac{S}{\sqrt{N-1}}}<1.81\)
• B. \(-183 < \frac{x}{\frac{S}{\sqrt{N-1}}}<1.83\)
• C. \(-137 < \frac{x}{\frac{S}{\sqrt{N-1}}}<1.37\)
• D. \(-2.23 < \frac{x}{\frac{S}{\sqrt{N-1}}}<2.23\)

Answer 1

The Correct Option is B

Step 1: Define the null hypothesis and test statistic. The null hypothesis is: \[ H_0: \mu = 0 \] The test statistic for the Student's \( t \)-test is: \[ t = \frac{x}{\frac{S}{\sqrt{N-1}}} \] 

Step 2: Determine the degrees of freedom. The degrees of freedom (\( D \)) for the test is: \[ D = N - 1 = 10 - 1 = 9 \] 

Step 3: Identify the confidence limits. At a 10\% significance level for a two-tailed test, the corresponding confidence level is \( 90\% \). From the table, for \( D = 9 \), the \( t \)-value is: \[ C(t) = \pm 1.83 \] 

Step 4: Formulate the test criterion. The null hypothesis will be accepted if: \[ -1.83 < \frac{x}{\frac{S}{\sqrt{N-1}}} < 1.83 \]