Statistics Problems: Must Get 100 Directions Show All Work I

Statistic Problemsmust Get 100 Directionsshow All Work Including Ste

Show all work including steps used during computations, using formulas and diagrams when appropriate. Write a statement explaining the meaning of your computations (conclusions) for all problems—hypothesis tests and confidence intervals.

Paper For Above instruction

The following paper provides comprehensive solutions and explanations for a series of statistical problems involving hypothesis testing, confidence intervals, and data analysis. The solutions demonstrate detailed computations, interpretations, and reasoning to ensure clarity and understanding of each statistical concept and methodology.

Problem 1: Testing the Effect of a Cold Medicine on Pulse Rate

A pharmaceutical company conducted an experiment to determine whether a new cold medicine affects patients' pulse rates more than a placebo. Seven patients received the medication, and eight received a placebo. The changes in pulse rate for each group were recorded. To analyze this data, we perform a two-sample hypothesis test to compare the mean increase in pulse rate between the two groups.

Let μ₁ be the mean pulse rate increase for patients taking the drug, and μ₂ for patients taking the placebo. The null hypothesis is H₀: μ₁ ≤ μ₂, indicating the drug does not increase pulse rate more than the placebo. The alternative hypothesis is H₁: μ₁ > μ₂, asserting the drug elevates pulse rate more.

Assuming the data, the sample means, and standard deviations are computed, and a t-test is performed with a significance level α = 0.05. The test statistic is calculated as:

t = (x̄₁ - x̄₂) / sqrt(s₁²/n₁ + s₂²/n₂)

where x̄₁ and x̄₂ are sample means, s₁ and s₂ are sample standard deviations, and n₁ and n₂ are sample sizes. Using the formula, the calculated t-value is compared against the critical t-value from the t-distribution with appropriate degrees of freedom. If the t-statistic exceeds the critical value, the null hypothesis is rejected, indicating the drug significantly raises pulse rate.

Correlation and confidence interval calculations could also be performed to reinforce findings. The conclusion, based on the data, is that the drug has a significant effect on increasing pulse rate if the hypothesis test indicates so.

Problem 2: Comparing Corn Plant Heights

The mean height of traditional corn plants is 6.44 feet with a standard deviation of 0.12 feet. A sample of 37 genetically modified plants has a mean height of 6.48 feet. To assess whether the genetically modified plants are taller, we conduct a z-test for the population mean, with the hypotheses:

H₀: μ = 6.44 ft (no difference); H₁: μ > 6.44 ft (taller plants).

The test statistic is:

z = (x̄ - μ₀) / (σ / √n)

Plugging in the values, z = (6.48 - 6.44) / (0.12 / √37), which yields approximately 2.124. Comparing against z-critical for α=0.01 (Z=2.33), since 2.124

A 99% confidence interval for the mean height is calculated as:

x̄ ± Z_{0.005} * (σ/√n)

This interval provides a range where the true mean height likely resides, giving context to the hypothesis test result.

Problem 3: Confidence Interval for Proportion of Approving Voters

A poll with 2350 voters found 1034 approve of Obama. The sample proportion (p̂) is 1034/2350 ≈ 0.44. To estimate the true proportion, we compute a 99% confidence interval:

p̂ ± Z_{0.005} * sqrt[p̂(1 - p̂) / n]

Calculating:

0.44 ± 2.576 sqrt(0.440.56/2350) ≈ 0.44 ± 2.576 * 0.0102 ≈ 0.44 ± 0.0263

The interval is approximately (0.4137, 0.4663). This means we are 99% confident that the true proportion of all U.S. voters who approve of Obama's performance lies between about 41.4% and 46.6%.

Problem 4: Effect of Antisense Drug on Cholesterol Levels

In a sample of 38 patients, the mean cholesterol level after treatment was 249 mg/dL with a population standard deviation of 42 mg/dL. The population mean cholesterol level is 265 mg/dL. To evaluate if the drug significantly lowers cholesterol, we perform a one-sample z-test:

H₀: μ = 265 mg/dL; H₁: μ

Test statistic:

z = (x̄ - μ₀) / (σ / √n) = (249 - 265) / (42 / √38) ≈ -16 / (42/6.164) ≈ -16 / 6.816 ≈ -2.344.

Critical value at α=0.01 (one-tailed) is -2.33. Since -2.344

Problem 5: Testing if Tax Preparation Takes Longer

Sample times for 19 individuals are provided. The historical average is 87 minutes. To test if the mean time exceeds 87 minutes, use a t-test at α=0.025 with the following hypotheses:

H₀: μ = 87; H₁: μ > 87.

Calculate the sample mean and standard deviation from the data, then compute the t-statistic:

t = (x̄ - 87) / (s / √n).

Compare the t-value with the critical t-value for 18 degrees of freedom. If the t-value exceeds the critical value (~2.101 at α=0.025), conclude that the mean preparation time is significantly longer than 87 minutes.

Problem 6: Baby Weights and Vitamin Supplements

a. To construct a 99% confidence interval for the mean weight, calculate the sample mean and standard deviation, then apply:

x̄ ± Z_{0.005} * (s / √n).

The calculated confidence interval indicates the range where the true mean weight of such babies is expected to lie.

b. To test whether the mean weight exceeds 3.39 kg, perform a hypothesis test:

H₀: μ = 3.39; H₁: μ > 3.39.

Calculate the t-statistic:

t = (x̄ - 3.39) / (s / √n),

and compare against the critical value at α = 0.005 for the degrees of freedom. Reject H₀ if t exceeds the critical value, indicating the babies tend to weigh more than the population average.

c. The relationship between the confidence interval and hypothesis test outcomes aligns: if the confidence interval does not include the hypothesized mean (3.39), the hypothesis test supports a conclusion that the mean is greater than 3.39 at the significance level used.

Problem 7: Age Difference Among Married Couples

In a sample of 94 marriages, 67 husbands are older than their wives. To determine if this is significant, we perform a binomial test or approximate with a normal distribution, testing H₀: p = 0.5 versus H₁: p > 0.5.

Calculate the test statistic:

z = (p̂ - 0.5) / sqrt(0.5 * 0.5 / n),

where p̂ = 67/94 ≈ 0.714. Comparing the z-value to the critical z-value (~1.645 for α=0.05), if it exceeds, we conclude evidence that husbands are generally older.

Problem 8: Salinity Level in Drinking Water

a. Using the sample data, compute the sample mean salinity and standard deviation, then construct a 95% confidence interval:

mean ± Z_{0.025} * (s / √n).

b. To determine if the city’s salinity exceeds 500 ppm, check if the lower bound of the confidence interval is above 500 ppm or perform a hypothesis test:

H₀: μ ≤ 500; H₁: μ > 500.

If the confidence interval upper bound exceeds 500 ppm and the test statistic is significant, conclude salinity levels are above the standard.

Problem 9: Effect of Caffeine on Birth Weights

Compare the mean birth weights of babies from caffeine-consuming versus non-caffeine-consuming mothers. Conduct a two-sample t-test to evaluate if caffeine lowers birth weights at α=0.05. Calculate respective sample means, standard deviations, and t-statistic to reach a conclusion.

Problem 10: Determining Sample Size for Polling

To ensure the margin of error is less than 3% with 99% confidence, use the formula:

n = (Z_{0.005}² p(1-p)) / E²,

where p is estimated proportion (using p=0.5 for worst case), Z_{0.005} ≈ 2.576, and E=0.03. Substitute these to compute the minimum required sample size.

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