Sample Of 25 Items Yields X Bar 60 Grams And S 9 Grams

4 A Sample Of 25 Items Yields X Bar 60 Grams And S 9 Grams Ass

Construct a 99 percent confidence interval for the population mean weight based on a sample of 25 items with a sample mean (X̄) of 60 grams and a sample standard deviation (s) of 9 grams, assuming a normal parent distribution.

Construct a 99 percent confidence interval for the percentage of trucks that had bad signal lights based on a sample of 900 trucks, of which 360 had bad signal lights.

Calculate the probability that a sample of 1600 voters in a region where 80 percent support a candidate will yield a sample proportion within 1 percentage point of the actual proportion.

Paper For Above instruction

The task involves multiple statistical analyses, including constructing confidence intervals for both a population mean and proportion, as well as calculating the probability associated with a sample proportion. Each part assumes different sampling conditions and distributions, which require appropriate statistical formulas and understanding.

Part 1: Confidence Interval for the Population Mean

The first analysis is constructing a 99% confidence interval for the population mean weight based on a sample of 25 items. Given the sample mean (X̄) of 60 grams, sample standard deviation (s) of 9 grams, and sample size (n) of 25, the approach involves using the t-distribution since the population standard deviation is unknown, and the sample size is relatively small.

The formula for the confidence interval for the mean is:

CI = X̄ ± t_{α/2, df} * (s / √n)

Where:

• X̄ = 60 grams
• s = 9 grams
• n = 25
• df = n - 1 = 24

Using a t-table or calculator for a 99% confidence level and 24 degrees of freedom, t_α/2 ≈ 2.796.

Calculating the standard error (SE):

SE = s / √n = 9 / √25 = 9 / 5 = 1.8

Constructing the interval:

Lower bound = 60 - 2.796 * 1.8 ≈ 60 - 5.033 = 54.967 grams

Upper bound = 60 + 2.796 * 1.8 ≈ 60 + 5.033 = 65.033 grams

Therefore, the 99% confidence interval for the population mean weight is approximately (54.97 grams, 65.03 grams).

Part 2: Confidence Interval for the Population Proportion

Next, for the truck signal lights problem, the sample size is 900 trucks, with 360 having bad signal lights. To estimate the proportion of trucks with bad lights with 99% confidence, we use the confidence interval for a proportion:

CI = p̂ ± z_α/2 * √[ p̂ (1 - p̂) / n ]

Where:

• p̂ = x / n = 360 / 900 = 0.4
• n = 900
• For 99% confidence, z_α/2 ≈ 2.576

Calculating standard error (SE):

SE = √[ p̂ (1 - p̂) / n ] = √[ 0.4 * 0.6 / 900 ] ≈ √[ 0.24 / 900 ] ≈ √0.0002667 ≈ 0.01633

Constructing the interval:

Lower bound = 0.4 - 2.576 * 0.01633 ≈ 0.4 - 0.0422 ≈ 0.3578

Upper bound = 0.4 + 2.576 * 0.01633 ≈ 0.4 + 0.0422 ≈ 0.4422

Thus, the 99% confidence interval for the true proportion of trucks with bad signal lights is approximately (0.3578, 0.4422), or (35.78%, 44.22%).

Part 3: Probability of Proportion Within a Margin

Finally, considering voters with an 80% support rate, we want the probability that a sample of 1600 voters will yield a sample proportion within 1 percentage point (0.01) of the true proportion.

Let p = 0.8, n = 1600, and the margin of error (E) = 0.01.

The sampling distribution of the sample proportion p̂ is approximately normal for large n, with mean p and standard deviation:

σ_p̂ = √[ p (1 - p) / n ] = √[ 0.8 * 0.2 / 1600 ] = √[ 0.16 / 1600 ] = √0.0001 = 0.01

The probability that p̂ is within 0.01 of 0.8 is:

P(0.79 ≤ p̂ ≤ 0.81) = P( (0.79 - p) / σ_p̂ ≤ Z ≤ (0.81 - p) / σ_p̂ )

Calculating Z-scores:

Z_lower = (0.79 - 0.8) / 0.01 = -0.01 / 0.01 = -1

Z_upper = (0.81 - 0.8) / 0.01 = 0.01 / 0.01 = 1

Using standard normal distribution tables or calculator, the probability becomes:

P( Z between -1 and 1 ) = 2 P( Z ≤ 1 ) - 1 ≈ 2 0.8413 - 1 = 0.6826

Hence, there is approximately a 68.26% probability that the sample proportion will lie within 1 percentage point of the true proportion, given the sample size and true proportion.

Conclusion

Through the above calculations, we have demonstrated how to construct confidence intervals for a population mean and proportion under specified confidence levels, and how to calculate the probability of observing a sample proportion within a specified margin of the true proportion using normal approximation. These techniques are fundamental tools in inferential statistics, supporting decision-making and estimation in various practical contexts such as quality control, survey analysis, and polling.

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