Statistics & Applications

Inferential

Medium

applied

A 95% confidence interval for a population mean was reported to be 152 to 160. If standard deviation $\sigma = 15$ . Then the sample size is : ( $Z_{0.025}$ =1.96)

✅ Correct Option: 3

Given information:

Confidence Interval: 152 to 160
Standard deviation $\sigma = 15$
$Z_{0.025} = 1.96$

The sample mean is the midpoint of the confidence interval.

$\bar{x} = \dfrac{152 + 160}{2}$

$\bar{x} = \dfrac{312}{2}$

$\bar{x} = 156$

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

$E = \dfrac{160 - 152}{2}$

$E = \dfrac{8}{2}$

$E = 4$

The margin of error formula for a confidence interval is:

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

Substituting the known values:

$4 = 1.96 \times \dfrac{15}{\sqrt{n}}$

$4 = \dfrac{29.4}{\sqrt{n}}$

$\sqrt{n} = \dfrac{29.4}{4}$

$\sqrt{n} = 7.35$

$n = (7.35)^2$

$n = 54.0225$

$n \approx 54$

Therefore, the sample size is 54.

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In a survey question for a sample of 340 people, 85 persons gave positive response, 145 people gave negative response and 110 people were neutral, gave no response. What is the point estimate of the proportion in the population who respond positive?

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Let us suppose that two independent random samples of sizes $n_1$ and $n_2$ has been drawn from the same normal population then degree of freedom of statistic t-distribution is:

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Consider the following hypothesis test,

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Match the LIST-I with LIST-II

LIST-I	LIST-II
A. Confidence interval	I. Sampling distribution of the sample mean approaches normal distribution as the sample size gets larger, no matter what the shape of the population distribution is.
B. Hypothesis	II. A measurable characteristic of a sample.
C. Central limit theorem	III. The amount of uncertainity associated with a sample estimate of a population parameter.
D. Statistic	IV. An educated guess which needs to be tested.

Choose the correct answer from the options given below:

2025: 22 May Shift 1

If n is the sample size of the population, then degree of freedom in t-distribution is

A 95% confidence interval for a population mean was reported to be 152 to 160. If standard deviation σ=15\sigma = 15σ=15 . Then the sample size is : (Z0.025Z_{0.025}Z0.025​=1.96)

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A 95% confidence interval for a population mean was reported to be 152 to 160. If standard deviation $\sigma = 15$ . Then the sample size is : ( $Z_{0.025}$ =1.96)