If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation = 50, then the size of the sample in the study is: (Given z_0.025 = 1.96) Option 1: 90 Option 2: 95 Option 3: 50 Option 4: 49

Answer: 49. Solution: The confidence interval is 132 to 160, with standard deviation = 50 and confidence level 95% using z_0.025 = 1.96. The sample mean is at the center of the confidence interval: x = (132 + 160)/2 x = 292/2 x = 146 The margin of error is the distance from the mean to either endpoint: E = (160 - 132)/2 E = 28/2 E = 14 The margin of error formula relates E, z, and n: E = z x ()/√n 14 = 1.96 x 50/√n 14 = 98/√n 14√n = 98 √n = 98/14 √n = 7 n = 49 Therefore, the sample size is n = 49.

Statistics & Applications

Inferential

Medium

applied

If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation $\sigma = 50$ , then the size of the sample in the study is:

(Given $z_{0.025} = 1.96$ )

✅ Correct Option: 4

The confidence interval is 132 to 160, with standard deviation $\sigma = 50$ and confidence level 95% using $z_{0.025} = 1.96$ .

The sample mean is at the center of the confidence interval:

$\bar{x} = \dfrac{132 + 160}{2}$

$\bar{x} = \dfrac{292}{2}$

$\bar{x} = 146$

The margin of error is the distance from the mean to either endpoint:

$E = \dfrac{160 - 132}{2}$

$E = \dfrac{28}{2}$

$E = 14$

The margin of error formula relates $E$ , $z$ , $\sigma$ , and $n$ :

$E = z \times \dfrac{\sigma}{\sqrt{n}}$

$14 = 1.96 \times \dfrac{50}{\sqrt{n}}$

$14 = \dfrac{98}{\sqrt{n}}$

$14\sqrt{n} = 98$

$\sqrt{n} = \dfrac{98}{14}$

$\sqrt{n} = 7$

$n = 49$

Therefore, the sample size is $n = 49$ .

2025: 26 May Shift 2

Consider the following test:

H₀: μ ≤ 12

H₁: μ > 12

A sample of 36 provided a sample mean $\bar{x} = 16$ and a sample standard deviation S=4.2. Then the value of the t- test statistic is:

2025: 22 May Shift 2

Which of the following are correct about t-test statistics?

(A) The mean of t-test distribution is 0.

(B) It depends on the degrees of freedom.

(D) With more degrees of freedom, it closely resembles standard normal distribution.

Choose the correct answer from the options given below:

2026: 25 May Shift 1

A simple random sample of 49 items from a population with standard deviation 10 resulted in a sample mean of 25. Then the 95% confidence interval for population mean are :

(Given $z_{\alpha/2} = 1.96$ where $\alpha = 0.05$ )

2025: 22 May Shift 2

A 95% confidence interval for a population mean was reported to be 152 to 160. If sample standard deviation $\sigma=15$ , then sample size used in this study is (Given $Z_{0.025}=1.96$ )

2025: 26 May Shift 2

With reference to sampling, which of the following are correct?

(A) Simple random sampling is probability sampling

(B) Snow-ball sampling is non-probability sampling

(D) Cluster sampling is non-probability sampling

Choose the correct answer from the options given below:

2026: 30 May Shift 1

In reference to the Inferential Statistics, which of the following statements are correct?

A. The number of degrees of freedom of a statistic is the number of independent varieties used to compute the statistic.

B. A confidence interval is a range that could be expected to contain the population parameter of interest.

C. In a hypothesis test, the significance level is the probability of making wrong decision when the null hypothesis is true.

D. The confidence level of the sampling distribution of a statistic is known as its standard error.

Choose the correct answer from the options given below:

If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation σ=50\sigma = 50σ=50, then the size of the sample in the study is: (Given z0.025=1.96z_{0.025} = 1.96z0.025​=1.96)

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If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation $\sigma = 50$ , then the size of the sample in the study is:

(Given $z_{0.025} = 1.96$ )