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).

Q: The probability of rejecting the null hypothesis when it is true, is called :

The probability of rejecting a true null hypothesis is the level of significance.Rejecting a null hypothesis when it is true is a Type I error, and the maximum probability of making it that a test allows is the level of significance, α, commonly set at 5% or 1%.The level of confidence is 1 − α, the probability of correctly retaining a true null hypothesis. The power of the test is 1 − β, the probability of correctly rejecting a false null hypothesis. Since these are three different quantities, 'all of the above' is also wrong.

Q: The following are percentages of fat found in 5 samples of each of two brands of food :A : 5.7, 4.5, 6.2, 6.3, 7.3B : 6.3, 5.7, 5.9, 6.4, 5.1To test the hypothesis of equal average fat content in the two types of food. Which of the following procedures is appropriate ?

The appropriate test is a two sample t-test with 8 degrees of freedom.The two sets of five samples come from two different brands of food. There is no natural pairing between the first sample of brand A and the first of brand B, so the samples are independent, and a paired t-test is not appropriate. With small samples and unknown variances, the means are compared by the two sample (pooled) t-test, whose degrees of freedom are:\[ \begin{aligned} df &= n_{1} + n_{2} - 2 \\ &= 5 + 5 - 2 = 8 \end{aligned} \]9 degrees of freedom subtracts only one. The paired options would be right only if each A value were matched with a B value, as in before-and-after measurements on the same units; a paired test on 5 pairs would have 4 degrees of freedom.

Q: In statistical hypothesis of equality of means such as H0 : μ = 10, if α = 5% :

With α = 5%, 5% of the time we will say that there is a real difference when there is no difference.α is the probability of a Type I error: rejecting H0 when it is true. Here H0 says the mean is 10, so rejecting it amounts to declaring a real difference from 10. If H0 is in fact true and the test is repeated many times, it will wrongly declare a difference in about 5% of cases.Saying there is no real difference when there is none is the correct decision, which happens 95% of the time, not 5%. An incorrect inference 95% of the time reverses the figures, and α says nothing about how often the null hypothesis is true, which rules out the last option.

Q: If one wishes to compare the means of two independent samples, the appropriate test statistics would be :

The appropriate test is the Student t-test.The independent-samples t-test compares the means of two independent groups, using the difference between the means relative to its standard error and the t-distribution, which suits the small samples common in biology.The F-test compares variances, or several means at once in ANOVA. The chi-square test compares observed and expected frequencies of categories, not means. Correlation measures the association between two variables measured on the same individuals.

Q: The probability of type one error is :

The probability of a type one error is the probability of rejecting the null hypothesis when it is true.A type I error is a false rejection: the null hypothesis is actually true, but the sample leads us to reject it (and accept the alternative). Its probability is the significance level \(\alpha\) chosen for the test.Accepting a false null hypothesis (B) is a type II error, with probability \(\beta\). Rejecting a false null (C) is a correct decision, the power of the test, and accepting a true null (D) is also a correct decision.

Q: If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample of size n from a normal population with mean \(\mu\) and variance \(\sigma^{2}\). What is the simple hypothesis of the following ?

The simple hypothesis is \(\mu = \mu_{0},\ \sigma^{2} = \sigma_{0}^{2}\).A hypothesis is simple when it specifies the population distribution completely. A normal distribution is fixed by its mean and variance, so stating both exactly, \(\mu = \mu_{0}\) and \(\sigma^{2} = \sigma_{0}^{2}\), leaves nothing unknown.The other options each contain an inequality (\(\mu \lt \mu_{0}\), \(\sigma^{2} \lt \sigma_{0}^{2}\) or \(\sigma^{2} \gt \sigma_{0}^{2}\)), which allows a whole range of distributions. Such hypotheses are composite.

Q: The P-value in hypothesis testing represents which of the following :

The P-value is the probability of observing results as extreme or more extreme than currently observed, given that the null hypothesis is true.It assumes the null hypothesis and asks how surprising the data are under it. A small P-value, below the chosen significance level such as 0.05, means such extreme results would rarely occur if the null hypothesis were true, so it is rejected.The common misreadings are the other options. The P-value is not the probability that the null hypothesis is true, because it is computed assuming that it is true. It is not the probability of failing to reject the null, and it is not the probability that the results 'are significant'; significance is a decision made by comparing the P-value with the chosen level.

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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The probability of rejecting the null hypothesis when it is true, is called :

A.Level of confidence
B.Level of significance
C.Power of the test
D.All of the above

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The probability of rejecting a true null hypothesis is the level of significance.

Rejecting a null hypothesis when it is true is a Type I error, and the maximum probability of making it that a test allows is the level of significance, α, commonly set at 5% or 1%.

The level of confidence is 1 − α, the probability of correctly retaining a true null hypothesis. The power of the test is 1 − β, the probability of correctly rejecting a false null hypothesis. Since these are three different quantities, 'all of the above' is also wrong.

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The following are percentages of fat found in 5 samples of each of two brands of food :

A : 5.7, 4.5, 6.2, 6.3, 7.3
B : 6.3, 5.7, 5.9, 6.4, 5.1

To test the hypothesis of equal average fat content in the two types of food. Which of the following procedures is appropriate ?

A.Two sample t-test with 8 degrees of freedom
B.Paired t-test with 4 degrees of freedom
C.Two sample t-test with 9 degrees of freedom
D.Paired t-test with 5 degrees of freedom

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The appropriate test is a two sample t-test with 8 degrees of freedom.

The two sets of five samples come from two different brands of food. There is no natural pairing between the first sample of brand A and the first of brand B, so the samples are independent, and a paired t-test is not appropriate. With small samples and unknown variances, the means are compared by the two sample (pooled) t-test, whose degrees of freedom are:

\[ \begin{aligned} df &= n_{1} + n_{2} - 2 \\ &= 5 + 5 - 2 = 8 \end{aligned} \]

9 degrees of freedom subtracts only one. The paired options would be right only if each A value were matched with a B value, as in before-and-after measurements on the same units; a paired test on 5 pairs would have 4 degrees of freedom.

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In statistical hypothesis of equality of means such as H₀ : μ = 10, if α = 5% :

A.95% of the time we will make an incorrect inference
B.5% of the time we will say that there is no real difference when there is no difference
C.5% of the time we will say that there is a real difference when there is no difference
D.95% of the time the null hypothesis will be correct

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

With α = 5%, 5% of the time we will say that there is a real difference when there is no difference.

α is the probability of a Type I error: rejecting H₀ when it is true. Here H₀ says the mean is 10, so rejecting it amounts to declaring a real difference from 10. If H₀ is in fact true and the test is repeated many times, it will wrongly declare a difference in about 5% of cases.

Saying there is no real difference when there is none is the correct decision, which happens 95% of the time, not 5%. An incorrect inference 95% of the time reverses the figures, and α says nothing about how often the null hypothesis is true, which rules out the last option.

Discuss in Forum

If one wishes to compare the means of two independent samples, the appropriate test statistics would be :

A.F-test
B.Student t-test
C.Chi-square
D.Correlation coefficient analysis

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The appropriate test is the Student t-test.

The independent-samples t-test compares the means of two independent groups, using the difference between the means relative to its standard error and the t-distribution, which suits the small samples common in biology.

The F-test compares variances, or several means at once in ANOVA. The chi-square test compares observed and expected frequencies of categories, not means. Correlation measures the association between two variables measured on the same individuals.

Discuss in Forum

The probability of type one error is :

A.Probability of rejecting null hypothesis (accepting alternative hypothesis) when null hypothesis is true
B.Probability of accepting null hypothesis (rejecting alternative hypothesis) when null hypothesis is false
C.Probability of rejecting null hypothesis (accepting alternative hypothesis) when null hypothesis is false
D.Probability of accepting null hypothesis (rejecting) alternative hypothesis) when null hypothesis is true

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: A

The probability of a type one error is the probability of rejecting the null hypothesis when it is true.

A type I error is a false rejection: the null hypothesis is actually true, but the sample leads us to reject it (and accept the alternative). Its probability is the significance level \(\alpha\) chosen for the test.

Accepting a false null hypothesis (B) is a type II error, with probability \(\beta\). Rejecting a false null (C) is a correct decision, the power of the test, and accepting a true null (D) is also a correct decision.

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If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample of size n from a normal population with mean \(\mu\) and variance \(\sigma^{2}\). What is the simple hypothesis of the following ?

A.\(\mu \lt \mu_{0},\ \sigma^{2} = \sigma_{0}^{2}\)
B.\(\mu = \mu_{0},\ \sigma^{2} = \sigma_{0}^{2}\)
C.\(\mu = \mu_{0},\ \sigma^{2} \lt \sigma_{0}^{2}\)
D.\(\mu = \mu_{0},\ \sigma^{2} \gt \sigma_{0}^{2}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: B

The simple hypothesis is \(\mu = \mu_{0},\ \sigma^{2} = \sigma_{0}^{2}\).

A hypothesis is simple when it specifies the population distribution completely. A normal distribution is fixed by its mean and variance, so stating both exactly, \(\mu = \mu_{0}\) and \(\sigma^{2} = \sigma_{0}^{2}\), leaves nothing unknown.

The other options each contain an inequality (\(\mu \lt \mu_{0}\), \(\sigma^{2} \lt \sigma_{0}^{2}\) or \(\sigma^{2} \gt \sigma_{0}^{2}\)), which allows a whole range of distributions. Such hypotheses are composite.

Discuss in Forum

The P-value in hypothesis testing represents which of the following :

A.The probability of failing to reject the null hypothesis, given the observed results
B.The probability that the null hypothesis is true, given the observed results
C.The probability that the observed results are statistically significant, given that the null hypothesis is true
D.The probability of observing results as extreme or more extreme than currently observed, given that the null hypothesis is true

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The P-value is the probability of observing results as extreme or more extreme than currently observed, given that the null hypothesis is true.

It assumes the null hypothesis and asks how surprising the data are under it. A small P-value, below the chosen significance level such as 0.05, means such extreme results would rarely occur if the null hypothesis were true, so it is rejected.

The common misreadings are the other options. The P-value is not the probability that the null hypothesis is true, because it is computed assuming that it is true. It is not the probability of failing to reject the null, and it is not the probability that the results 'are significant'; significance is a decision made by comparing the P-value with the chosen level.

Discuss in Forum

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