Suppose there is a population of test scores on a large, standardized exam for which the mean and standard deviation are unknown. Two different random samples of 50 data values are taken from the population. One sample has a larger sample standard deviation (SD) than the other. Each of the samples is used to construct a 95% confidence interval. How do you think these two confidence intervals would compare?

The confidence interval based on the sample with the larger standard deviation would be wider.

Step-by-step explanation:

The confidence interval expresses the range of values in which the true mean can exist in with a certain level.of confidence.

It is usually calculated thus,

Confidence interval = (Sample mean) ± (Margin of error)

So, it is evident that the width of the range is determined by the Margin of Error.

Margin of error = (critical value) × (Standard error of the mean)

The critical value depends on solely the sample size, and the confidence level. This would be he same for the two distributions being considered. Hence, they would have the same critical value.

But Standard Error of the mean, σₓ, is given as

σₓ = (σ/√n)

σ = standard deviation

n = Sample size (equal for both data set)

This shows that the distribution with the higher standard deviation has a standard error of the mean.

This translates to a higher margin of error and hence, a wider confidence interval.

Hope this Helps!!!

2500

step-by-step explanation:

answer:   x*94 +688*6   = 267*100

94x+4128=26700    

94x=22572     /94

x=240.128

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