Round All Your Answers To 4 Decimal Places Part I

Round All Your Answers To 4 Decimal Places Part I Focusing On Chapte

Suppose 14% of people are left handed. What is the probability that a random sample of 200 people will have less than 12% lefties? Be sure to check the conditions to make the necessary assumptions before using the model. Conditions: Randomization, 10% Condition, Success/Failure Condition, Probability. The life of General Electric light bulbs are normally distributed with a mean of 200 hours and a standard deviation of 20 hours. What is the probability that a randomly selected light bulb will last more than 210 hours? What is the probability that a random sample of 10 light bulbs has a mean life greater than 210 hours? (check assumptions and conditions first) Conditions: Randomization, 10% Condition, Large Enough Sample Condition, Probability.

Paper For Above instruction

The statistical analysis of proportions and means involves understanding the underlying assumptions and applying appropriate models to determine probabilities and confidence intervals. In this context, we explore the likelihood of observing certain proportions within a sample, given a hypothesized population proportion, and assess whether sample data provide evidence for a claim or hypothesis.

Part I: Probability Related to Left-Handedness and Light Bulb Lifespan

First, consider the probability that a sample of 200 people contains fewer than 12% left-handed individuals when the true proportion of left-handed people in the population is 14%. To address this, we model the number of left-handed individuals as a binomial distribution, which can be approximated by a normal distribution under certain conditions. These conditions are: the sample should be random (to ensure independence), the sample size should be less than 10% of the population (to satisfy the 10% condition), and the sample should have enough successes and failures (both np and n(1-p) should be at least 10).

Calculating the mean and standard deviation for the sampling distribution of the proportion, we have:

θ = 0.14, n = 200

Mean: μ_p̂ = p = 0.14

Standard deviation: σ_p̂ = sqrt [ p(1 - p) / n ] = sqrt [ 0.14 * 0.86 / 200 ] ≈ 0.0222

We want to find P(p̂

z = (0.12 - 0.14) / 0.0222 ≈ -0.02 / 0.0222 ≈ -0.9009

Using standard normal distribution tables or software, P(z

Therefore, the probability that fewer than 12% of the sample are left-handed is approximately 0.1833.

Lighting Bulb Lifespan Probability Analysis

Next, considering the lifetime of General Electric light bulbs which is normally distributed with a mean (μ) of 200 hours and a standard deviation (σ) of 20 hours, we determine the probability that a randomly selected bulb lasts more than 210 hours. The condition for normal distribution applying is satisfied, given the population parameters are known.

The z-score for 210 hours:

z = (210 - 200) / 20 = 10 / 20 = 0.5

Using the standard normal table, P(Z > 0.5) = 1 - P(Z

Thus, the likelihood that a randomly selected bulb lasts more than 210 hours is approximately 0.3085.

Furthermore, for a sample of size 10 bulbs, we look for the probability that the mean lifespan exceeds 210 hours. The sampling distribution of the sample mean μ̄ has the same mean as the population (200 hours), but the standard error decreases:

SE = σ / sqrt(n) = 20 / sqrt(10) ≈ 6.3246

The z-score for the sample mean of 210 hours:

z = (210 - 200) / 6.3246 ≈ 10 / 6.3246 ≈ 1.5811

Probability that the sample mean exceeds 210 hours:

P(μ̄ > 210) = 1 - P(Z

Conditions checked include the randomness of sampling and the sample size being sufficiently large for the normal approximation to be valid, especially since the sample size is relatively small but the underlying distribution is normal, satisfying the conditions for inference about the mean.

Part II: Testing Belief in Ghosts in Young Adults

In 2005, 32% of adults believed in ghosts. A 2005 Gallup Poll sampled 200 young adults, finding 38% believed in ghosts. The question is whether this sample provides statistical evidence that young adults are more likely than the general adult population to believe in ghosts.

The population of interest is all young adults aged 18-29. The parameter p denotes the true proportion of young adults who believe in ghosts.

Our hypotheses for the test are:

  • Null hypothesis, H0: p = 0.32
  • Alternative hypothesis, HA: p > 0.32

The success count in the sample is 38% of 200, i.e., 0.38 * 200 = 76 individuals.

Calculations: In the sample, p̂ = 76 / 200 = 0.38

Check assumptions and conditions: the sample is random, less than 10% of the population (assuming the total population of young adults exceeds 2000), and np0 = 200 0.32 = 64 ≥ 10, n(1 - p0) = 200 0.68 = 136 ≥ 10, satisfying successes/failures condition.

Compute the test statistic (z):

z = (p̂ - p0) / sqrt[ p0(1 - p0) / n ] = (0.38 - 0.32) / sqrt[ 0.32 * 0.68 / 200 ]

Standard error: sqrt[ 0.2176 / 200 ] ≈ 0.0465

z ≈ 0.06 / 0.0465 ≈ 1.2903

Find the P-value: P(Z > 1.2903) ≈ 0.0980.

Conclusion: Since P-value ≈ 0.098, which is less than the significance level α = 0.10, we reject H0 and conclude there is statistically significant evidence at the 10% level that young adults are more likely to believe in ghosts.

Understanding Errors in Context

A Type I error involves rejecting a true null hypothesis. In this context, it would mean concluding that young adults are more likely to believe in ghosts when, in fact, the population proportion is still 32%. Conversely, a Type II error would involve failing to reject the null hypothesis when the alternative is true, implying we do not detect an increased belief when it actually exists.

Confidence Interval for p

Constructing a 95% confidence interval for p involves the formula:

p̂ ± Z_0.025 * SE, where Z_0.025 ≈ 1.96

Standard error: as above, SE ≈ 0.0465

Interval:

0.38 ± 1.96 * 0.0465 ≈ 0.38 ± 0.0911

So, the 95% confidence interval is approximately (0.2889, 0.4711).

This means we are 95% confident that the true proportion of young adults who believe in ghosts lies within this interval. Interpretation: "We are 95% confident that between approximately 28.89% and 47.11% of young adults believe in ghosts." Increasing the confidence level to 99% with a margin of error of only 0.03 would require increasing the sample size, calculated using the formula:

n = (Z * σ / E)^2;

where Z ≈ 2.576 for 99% confidence, σ ≈ √[p̂(1 - p̂)] ≈ sqrt(0.38 * 0.62) ≈ 0.487, and E = 0.03

Sample size n ≈ (2.576 * 0.487 / 0.03)^2 ≈ (39.52)^2 ≈ 1562.69, roughly 1563.

Part III: Car Weight Analysis

The population of interest is all licensed cars in the United States. The hypothesis test assesses whether the average weight differs from 3000 pounds.

The sample size is n = 80 with a sample mean of 2910 pounds and a standard deviation of 532 pounds. The null hypothesis: H0: μ = 3000. The alternative hypothesis: HA: μ ≠ 3000.

Check assumptions: the sample is random, and with n = 80, the Central Limit Theorem suggests the sampling distribution of the mean is approximately normal. The standard error:

SE = σ / sqrt(n) = 532 / sqrt(80) ≈ 59.55

Calculate the z-score:

z = (2910 - 3000) / 59.55 ≈ -1.529

Find the P-value for a two-tailed test:

P = 2 P(Z 0.0632 = 0.1264.

Since P ≈ 0.1264 > 0.05, we fail to reject H0 at the 5% significance level, suggesting insufficient evidence to conclude that the average weight differs from 3000 pounds. The 90% confidence interval is:

2910 ± Z_0.05 * 59.55, where Z_0.05 ≈ 1.645

Interval: 2910 ± 1.645 * 59.55 ≈ 2910 ± 98.02, which yields (2811.98, 3008.02).

This interval suggests we are 90% confident that the true average weight of all cars is between approximately 2812 and 3008 pounds. To increase confidence to 95% while reducing the margin of error to 50 pounds, the new sample size n can be estimated by:

n = (Z σ / E)^2 = (1.96 532 / 50)^2 ≈ (20.84)^2 ≈ 434.34, roughly 435 cars.

Conclusion

Robust statistical analysis involving hypothesis testing and confidence interval estimation enables researchers to draw meaningful inferences about population parameters while respecting assumptions and conditions. These methods are crucial tools for making data-driven decisions across various fields, including social sciences, engineering, and public policy.

References

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