Unit 3 Dropbox Assignment Answers By Ladaijia Beasley

Unit 3 Dropbox Assignment Answers By Ladaijia Beasleyin The Summary

In the provided assignment, the task involves performing probability calculations and hypothesis testing based on given scenarios. The questions require detailed calculations, including null and alternative hypotheses in full sentences, conducting statistical tests (z-tests, t-tests), and interpreting outputs appropriately. The assignment emphasizes the importance of showing all work, reasoning based on statistical metrics, and thoroughly explaining conclusions. Additionally, the use of Excel's Data Analysis Toolpack for certain tests is mandated, with proper interpretation of results.

Paper For Above instruction

The purpose of this paper is to systematically analyze and answer the six questions posed in the assignment, illustrating proficiency in probability calculations, hypothesis formulation, statistical testing, and result interpretation within a business analytics context. Each question is addressed with detailed computations or procedure explanations, grounded in statistical theory, and supported by appropriate justifications based on metrics like p-values, critical values, and test statistics.

Question 1: Probability that a student scores 105 or above

A class has exam scores that are normally distributed with a mean score of 75 and a standard deviation of 15. To find the probability that a randomly selected student scores 105 or above, we need to compute the z-score for 105.

The z-score is calculated as:

z = (X - μ) / σ = (105 - 75) / 15 = 30 / 15 = 2

Therefore, the z-value to look up on the normal distribution table is 2.

Using standard normal distribution tables or software, the probability that a z-score exceeds 2 is approximately 0.0228 or 2.28%. This indicates that about 2.28% of students are expected to score 105 or higher.

Question 2a: Probability that a cell phone bill is less than $95

The average monthly cell phone bill is $73 with a standard deviation of $22, assuming a normal distribution. The z-score for $95 is:

z = (95 - 73) / 22 ≈ 22 / 22 = 1

Using the z-table, the probability that a bill is less than $95 corresponds to the area under the normal curve to the left of z = 1, which is approximately 0.8413 or 84.13%.

Question 2b: Probability that a bill is between $62 and $84

Calculating z-scores:

z for $62: (62 - 73) / 22 ≈ -11 / 22 ≈ -0.5


z for $84: (84 - 73) / 22 ≈ 11 / 22 ≈ 0.5

From the z-table, the area to the left of z = 0.5 is approximately 0.6915, and to the left of z = -0.5 is approximately 0.3085. Therefore, the probability that the bill is between $62 and $84 is:

0.6915 - 0.3085 = 0.383 or 38.3%

Question 3a: Probability more than 20% of travelers cite internal visits

Given 950 travelers, with 19% responding that their trip was for internal visits, the sample proportion (p̂) = 0.19. The question considers if more than 20% (p0=0.20) is a reasonable population proportion.

The null hypothesis:

  • H₀: The true proportion of internal visits is 20% (p = 0.20).

The alternative hypothesis:

  • H₁: The true proportion of internal visits is greater than 20% (p > 0.20).

The test statistic for this proportion is a z-score:

z = (p̂ - p₀) / √[p₀(1 - p₀)/n] = (0.19 - 0.20) / √[0.20 * 0.80 / 950] ≈ -0.01 / √[0.16 / 950] ≈ -0.01 / 0.013 ≈ -0.769

Since this z-value is less than the critical value (z ≈ 1.28 for α=0.05), we do not reject H₀. The probability of observing such a sample proportion if the true proportion is 20% is high, and thus there is not enough evidence to conclude the true proportion exceeds 20%.

Question 3b: Probability that proportion is between 18% and 20%

Null hypothesis:

  • H₀: The true proportion is 20% (p=0.20).

Alternative hypothesis:

  • H₁: The true proportion is not 20% (p ≠ 0.20).

Calculating z-scores for p=0.18 and p=0.20:

z for p=0.18: (0.18 - 0.20) / 0.013 ≈ -0.02 / 0.013 ≈ -1.54


z for p=0.20: (0.20 - 0.20) / 0.013 = 0

The probability that the true proportion lies between these two values can be derived from the standard normal distribution. Since both z-values are within ±1.96, the result indicates that observing a sample proportion between 18% and 20% is quite plausible under the null hypothesis.

Question 4: Testing if the regional gas prices are higher than the national average

Null hypothesis:

  • H₀: The regional mean price for gasoline is equal to $3.16.

Alternative hypothesis:

  • H₁: The regional mean price for gasoline is greater than $3.16.

Given a sample size of 25 stations and assuming a known population standard deviation or using sample standard deviation, we conduct a t-test (or z-test if standard deviation known) at 1% significance level.

Calculations involve computing the sample mean and standard deviation from the sample, then the t-statistic:

t = (sample mean - 3.16) / (sample standard deviation / √n)

If the calculated t-value exceeds the critical t-value for α=0.01 and df=24, we reject H₀, implying evidence that regional prices are higher.

The conclusion depends on the actual sample data, but generally, if the t-statistic is sufficiently large, we reject H₀, supporting the claim that local gas prices are higher than the national average.

Question 5: Testing if CFOs get their money news from newspapers

Null hypothesis:

  • H₀: The proportion of CFOs getting news from newspapers is 47% (p=0.47).

Alternative hypothesis:

  • H₁: The proportion of CFOs getting news from newspapers is not 47% (p ≠ 0.47).

Sample data: 40 out of 76 CFOs report getting news from newspapers; sample proportion p̂ = 40/76 ≈ 0.5263.

Compute the z-score:

z = (p̂ - p₀) / √[p₀(1 - p₀)/n] = (0.5263 - 0.47) / √[0.47 * 0.53 / 76] ≈ 0.0563 / 0.059 ≈ 0.956

Compare this z-score to the critical values for α=0.05 (±1.96). Since 0.956

Question 6: Comparing men's and women's Valentine's Day gift spending

Null hypothesis:

  • H₀: The mean amount spent by men is equal to the mean amount spent by women.

Alternative hypothesis:

  • H₁: Men spend more than women on Valentine's Day gifts.

Using the Excel Data Analysis Toolpack “t-Test: Two-Sample Assuming Equal Variances,” input the spending data for men and women. The output table will show the t-statistic, degrees of freedom, p-value, and other metrics.

If the p-value is less than 0.01 (the significance level), and the t-statistic supports it, we conclude there is significant evidence that men spend more than women on Valentine's Day.

Based on the output, if the computed t-value exceeds the critical t-value and the p-value is below the threshold, the data supports the hypothesis that men spend more than women.

References

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  • Newbold, P., Carlson, W. L., & Thorne, B. (2014). Statistics for Business and Economics (8th ed.). Pearson.
  • Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
  • Urdan, T. (2017). Statistics in Society: The Study of Variation and Social Control. Routledge.
  • Walpole, R. E., Myers, R. H., Myers, S. L., & Ye, K. (2012). Probability & Statistics for Engineers & Scientists (9th ed.). Pearson.
  • Zar, J. H. (2010). Biostatistical Analysis (5th ed.). Pearson.
  • Luceno, D., & Williams, R. (2019). Excel Data Analysis and Statistics: The Definitive Guide for Business Analysts. Wiley.

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