Consider the following hypothesis test: H_0: = 18 H_1: ≠ 18 If a sample of 48 provided a sample mean x = 17 and a sample standard deviation = 4.5, then the value of the t-test statistic is: Option 1: 2.14 Option 2: -1.54 Option 3: 0.84 Option 4: 1.988

Answer: -1.54. Solution: The t-test formula is: t = (x - )/(s/√n) Given information: - Sample mean (x) = 17 - Population mean under H_0 () = 18 - Sample standard deviation (s) = 4.5 - Sample size (n) = 48 The standard error is: s/√n = 4.5/√48 = 4.5/6.928 = 0.6495 The t-statistic is: t = (x - )/(s/√n) = (17 - 18)/0.6495 = (-1)/0.6495 = -1.54 Therefore, the value of the t-test statistic is -1.54.

Statistics & Applications

Inferential

Medium

applied

Consider the following hypothesis test:

$H_0: \mu = 18$

$H_1: \mu \neq 18$

If a sample of 48 provided a sample mean $\bar{x} = 17$ and a sample standard deviation $\sigma = 4.5$ , then the value of the t-test statistic is:

2.14

-1.54

0.84

1.988

✅ Correct Option: 2

The t-test formula is:

$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$

Given information:

Sample mean ( $\bar{x}$ ) = 17
Population mean under $H_0$ ( $\mu$ ) = 18
Sample standard deviation ( $s$ ) = 4.5
Sample size ( $n$ ) = 48

The standard error is:

$\frac{s}{\sqrt{n}} = \frac{4.5}{\sqrt{48}}$

$= \frac{4.5}{6.928}$

$= 0.6495$

The t-statistic is:

$t = \frac{\bar{x} - \mu}{s/\sqrt{n}}$

$= \frac{17 - 18}{0.6495}$

$= \frac{-1}{0.6495}$

$= -1.54$

Therefore, the value of the t-test statistic is $-1.54$ .

2025: 13 May Shift 2

When the two independent small samples of sizes $n_1$ and $n_2$ with means $\bar{x}_1$ and $\bar{x}_2$ respectively are drawn from populations with identical population variances, the test-statistic is computed as

2025: 3 June Shift 2

Which of the following statements are TRUE?

(A) The variable t of t-distribution ranges from $-\infty$ to $\infty$ .

(B) The probability curve of the t-distribution is symmetric about the line $t=0$

(D) As the number of degrees of freedom decreases, the t-distribution curve moves closer to the standard normal probability curve.

Choose the correct answer from the options given below:

2025: 29 May Shift 2

If a sample has 'n' observation $x_1$ , $x_2$ ............. $x_n$ with 'm' constraints on these values then degree of freedom of sample statistic is:

2026: 30 May Shift 1

A company has been producing steel tubes of mean inner diameter of 2 cm. A sample of 10 tubes gives an inner diameter of 2.01 cm and a standard deviation of 0.063 cm. If $t_9(0.05)=2.262$ , then which of the following is NOT correct?

2025: 14 May Shift 1

Which of the following are the assumptions underlying the use of t-distribution?

(A) The variance of population is known.

(B) The samples are drawn from a normally distributed population.

(D) It depends on a parameter known as degree of freedom.

Choose the correct answer from the options given below:

2026: 22 May Shift 2

A company has been producing steel tubes of mean inner diameter of 3 cm. A sample of 17 tubes gives an inner diameter of 3.01 cm and a standard deviation of 0.064. Then the value of the t-test statistic is:

Consider the following hypothesis test: H0:μ=18H_0: \mu = 18H0​:μ=18 H1:μ≠18H_1: \mu \neq 18H1​:μ=18 If a sample of 48 provided a sample mean xˉ=17\bar{x} = 17xˉ=17 and a sample standard deviation σ=4.5\sigma = 4.5σ=4.5, then the value of the t-test statistic is:

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CUET Mathematics Past Year Questions (Topic-Wise)

Consider the following hypothesis test:

$H_0: \mu = 18$

$H_1: \mu \neq 18$

If a sample of 48 provided a sample mean $\bar{x} = 17$ and a sample standard deviation $\sigma = 4.5$ , then the value of the t-test statistic is: