Question

A 98% confidence interval for a proportion is found to be (0.62, 0.68).

What is
the sample proportion?

O A. 0.66

OB. 0.64

O C. 0.65

O D. 0.68

Answer 1

Answer:

I think the answer is C ,The sample proportion is the fraction of samples which were successes, so. (1) For large , has an approximately normal distribution.

Step-by-step explanation:

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Answer 2

Given that a 98% confidence interval for a proportion is found to be (0.62, 0.68). The sample proportion will be 0.65.

What is confidence interval for population proportion?

Formula to calculate the confidence interval for population proportion p :-

$p' - Z_{\alpha }\sqrt{\frac{p'q'}{n} } < p < p' + Z_{\alpha }\sqrt{\frac{p'q'}{n} }$

Where,

p is the population proportion.

p' is the sample proportion.

n is the size of the sample.

$Z_{\alpha }$ is the desired degree of confidence.

Given that a 98% confidence interval for a proportion p is found to be (0.62, 0.68).

$p' - Z_{\alpha }\sqrt{\frac{p'q'}{n} } = 0.62\\\\p' + Z_{\alpha }\sqrt{\frac{p'q'}{n} } = 0.68$

Adding the two equations,

2p' = 1.3

p' = 0.65

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