Hypothesis Testing Questions and Answers

Q: For accepting or rejecting a null hypothesis, which one of the following is NOT used as a significance test method in statistics?

The one NOT used as a significance test is relative standard deviation.The relative standard deviation (coefficient of variation) is a descriptive measure of spread, the standard deviation as a fraction of the mean. It has no test statistic or sampling distribution, so it cannot be used to accept or reject a null hypothesis.The Z-test and Student's t-test compare a statistic with a known distribution to judge significance. Pearson's correlation coefficient comes with a standard significance test (via a t-statistic) of the null hypothesis that the true correlation is zero.

Q: A simple random sample of size 10 from \(N(\mu, \sigma^{2})\) gives 98% confidence interval (20.49, 23.51). Then the null hypothesis \(H_{0} : \mu = 20.5\) against \(H_{A} : \mu \neq 20.5\)

The hypothesis can be rejected at 10% level of significance.The interval is centred at \(\bar{x} = 22\) with half-width 1.51. With \(n = 10\) and \(\sigma\) unknown, the 98% interval uses \(t_{0.01, 9} = 2.821\), so the standard error is:\[ \begin{aligned} s/\sqrt{n} &= 1.51/2.821 = 0.535 \\ t &= (22 - 20.5)/0.535 = 2.80 \end{aligned} \]Since 20.5 lies just inside the 98% interval, \(H_{0}\) cannot be rejected at 2%, so the first option is wrong. But \(2.80\) exceeds \(t_{0.025, 9} = 2.262\) and \(t_{0.05, 9} = 1.833\), so \(H_{0}\) is rejected at 5% and at 10%. That makes 'cannot be rejected at 5%' and 'cannot be rejected at any level' wrong. The same conclusion follows if \(\sigma\) is treated as known (z = 2.31 against 1.96 and 1.645).

For accepting or rejecting a null hypothesis, which one of the following is NOT used as a significance test method in statistics?

A.Z-test
B.Student’s t-test
C.Pearson’s correlation
D.Relative standard deviation

Exam Year(s)

GATE AG - 2018,

Answer: D

The one NOT used as a significance test is relative standard deviation.

The relative standard deviation (coefficient of variation) is a descriptive measure of spread, the standard deviation as a fraction of the mean. It has no test statistic or sampling distribution, so it cannot be used to accept or reject a null hypothesis.

The Z-test and Student's t-test compare a statistic with a known distribution to judge significance. Pearson's correlation coefficient comes with a standard significance test (via a t-statistic) of the null hypothesis that the true correlation is zero.

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A simple random sample of size 10 from \(N(\mu, \sigma^{2})\) gives 98% confidence interval (20.49, 23.51). Then the null hypothesis \(H_{0} : \mu = 20.5\) against \(H_{A} : \mu \neq 20.5\)

A.can be rejected at 2% level of significance
B.cannot be rejected at 5% level of significance
C.can be rejected at 10% level of significance
D.cannot be rejected at any level of significance

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

The hypothesis can be rejected at 10% level of significance.

The interval is centred at \(\bar{x} = 22\) with half-width 1.51. With \(n = 10\) and \(\sigma\) unknown, the 98% interval uses \(t_{0.01, 9} = 2.821\), so the standard error is:

\[ \begin{aligned} s/\sqrt{n} &= 1.51/2.821 = 0.535 \\ t &= (22 - 20.5)/0.535 = 2.80 \end{aligned} \]

Since 20.5 lies just inside the 98% interval, \(H_{0}\) cannot be rejected at 2%, so the first option is wrong. But \(2.80\) exceeds \(t_{0.025, 9} = 2.262\) and \(t_{0.05, 9} = 1.833\), so \(H_{0}\) is rejected at 5% and at 10%. That makes 'cannot be rejected at 5%' and 'cannot be rejected at any level' wrong. The same conclusion follows if \(\sigma\) is treated as known (z = 2.31 against 1.96 and 1.645).

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