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let x, the weight of a $2 fish fry at a local restaurant follows normal distribution. manager wants to be sure that the variance of the weight of fish fry does not exceed 0.05. derive a required formula for one sided confidence interval, and hence calculate a 95% confidence interval based upon a sample of size 10 with sample variance 0.014. does this sample support the manager ?

User Amruth
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95% one-sided upper confidence interval for the population variance is (0, 0.0235].

This means that we are 95% confident that the true population variance of fish fry weight is less than or equal to 0.0235.

Since the calculated upper bound of the confidence interval (0.0235) is less than the manager's desired maximum variance of 0.05, the sample supports the manager's belief that the variance of fish fry weight is indeed within acceptable limits.

Formula for one-sided upper confidence interval for variance:

The formula for a one-sided upper confidence interval for variance (σ²) with confidence level 1 - α (e.g., 95%) is:

σ² ≤ s² + [(n - 1) / χ²(α, n - 1)] * s²

where:

σ² is the population variance

s² is the sample variance (0.014 in this case)

n is the sample size (10)

χ²(α, n - 1) is the upper χ² critical value for α and n - 1 degrees of freedom. For a 95% confidence level and 9 degrees of freedom (n - 1), the critical value is approximately 14.684.

Calculating the confidence interval:

Plugging in the values:

σ² ≤ 0.014 + [(10 - 1) / 14.684] * 0.014 ≈ 0.014 + 0.0095 ≈ 0.0235

Therefore, the 95% one-sided upper confidence interval for the population variance is (0, 0.0235].

Interpretation:

This means that we are 95% confident that the true population variance of fish fry weight is less than or equal to 0.0235.

Since the calculated upper bound of the confidence interval (0.0235) is less than the manager's desired maximum variance of 0.05, the sample supports the manager's belief that the variance of fish fry weight is indeed within acceptable limits.

User Ricka
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