Hypothesis Testing Questions and Answers

Q: ................... means rejection of hypothesis which should have been accepted.

Rejecting a hypothesis that should have been accepted is a Type-I error.A Type I error occurs when a true null hypothesis is rejected; its probability is the level of significance, alpha. A Type II error is the reverse: accepting (failing to reject) a null hypothesis that is false.The null and alternative hypotheses are the statements being tested, not errors in deciding between them.

Q: A random sample of size 26 has a mean of 20. The sum of squares of the deviations taken from the mean is 200. If the population mean is 18, what is the value of t-statistic ?

~ 3.6The sample variance is \[ s^{2} = \frac{200}{26-1} = 8,\quad s = 2.83 \] \[ t = \frac{\bar{x}-\mu}{s/\sqrt{n}} \] \[ t = \frac{20-18}{2.83/\sqrt{26}} \approx 3.6 \] So the t-statistic is about 3.6. The other values come from dividing by n or by a wrong standard error.The other options, ~ 0.9; ~ 2.1; ~ 5.2, are not the answer to this question. This agrees with the official final answer key for the paper, which gives option C.

11.

................... means rejection of hypothesis which should have been accepted.

A.Alternative Hypothesis
B.Null Hypothesis
C.Type-II error
D.Type-I error

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

Rejecting a hypothesis that should have been accepted is a Type-I error.

A Type I error occurs when a true null hypothesis is rejected; its probability is the level of significance, alpha. A Type II error is the reverse: accepting (failing to reject) a null hypothesis that is false.

The null and alternative hypotheses are the statements being tested, not errors in deciding between them.

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12.

The following statistical test is used to ascertain whether there is significant difference between the variances of two sets of observations :

A.t-test
B.F-test
C.Chi square test
D.Regression

Exam Year(s)

UGC NET - 2012,

Answer: B

The F-test is used to find out whether there is a significant difference between the variances of two sets of observations. It takes the ratio of the two sample variances and compares it with the tabulated F value for the degrees of freedom. The t-test compares means, the chi-square test examines goodness of fit or independence between categorical data, and regression relates a dependent variable to independent variables. Neither of these tests the equality of variances.

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13.

Two normal populations have variances \(\sigma_1^{2}\) = 10 and \(\sigma_2^{2}\) = 20. Two random samples of sizes 25 and 20, independently selected from these populations have variances of \(S_1^{2}\) = 8 and \(S_2^{2}\) = 15, respectively. What is the \(F_{(24, 19)}\) statistic ?

A.1
B.2
C.2.81
D.3.6

Exam Year(s)

UGC NET - 2013,

Answer: A

The F statistic is the ratio of the sample variances each divided by its population variance.

\[ F = \frac{S_1^2 / \sigma_1^2}{S_2^2 / \sigma_2^2} \]

\[ F = \frac{8/10}{15/20} \]

\[ F = \frac{0.8}{0.75} \approx 1.07 \]

The nearest value in the options is 1, option A, which is the keyed answer. The values 2, 2.81 and 3.6 are much larger.

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14.

A random sample of size 26 has a mean of 20. The sum of squares of the deviations taken from the mean is 200. If the population mean is 18, what is the value of t-statistic ?

A.~ 0.9
B.~ 2.1
C.~ 3.6
D.~ 5.2

Exam Year(s)

UGC NET - 2013,

Answer: C

~ 3.6

The sample variance is \[ s^{2} = \frac{200}{26-1} = 8,\quad s = 2.83 \] \[ t = \frac{\bar{x}-\mu}{s/\sqrt{n}} \] \[ t = \frac{20-18}{2.83/\sqrt{26}} \approx 3.6 \] So the t-statistic is about 3.6. The other values come from dividing by n or by a wrong standard error.

The other options, ~ 0.9; ~ 2.1; ~ 5.2, are not the answer to this question. This agrees with the official final answer key for the paper, which gives option C.

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15.

A sample size of 17 observations is selected from a normal population with mean = 50. The sample mean and variance are 48 and 8 respectively. The value of t - statistic is

A.0.25
B.- 2.82
C.- 2.2
D.0.71

Exam Year(s)

UGC NET - 2014,

Answer: B

The t statistic is the difference between the sample mean and the population mean divided by the standard error of the mean. With n = 17 and an unbiased estimate of variance from the given figure, the value is close to -2.82.

\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]\[ = \frac{48 - 50}{\sqrt{8.5/17}} \]\[ = \frac{-2}{0.707} \approx -2.82 \]

0.25 and 0.71 are of the wrong sign, and -2.2 does not follow from the data.

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