Assignment 31: How To Determine The Average Number Of Hours

Assignment 31you Want To Determine The Average Number Of Hours Worked

ASSIGNMENT .You want to determine the average number of hours worked by professors at your university per week. You want to be able to express your answer as a confidence interval, with 95% confidence (that is, use z = 1.96) and you are aiming for a precision of plus or minus two hours. Past studies of this type suggest that the variance in professor work hours is about 90 (that is, σ² = 90). a. How big is the minimum required sample size that would allow you to meet your confidence and precision goals? b. If you wanted your precision to double (that is, your precision would now be plus or minus one hour), how does that affect your minimum required sample size? c. (Reset precision to the original setting.) If you wanted your confidence to increase to 99% (that is, use z = 2.575 instead of 1.96), how does that affect your minimum required sample size? A publishing house conducted a survey to assess the reading habits of teenagers. The company publishes four types of books specifically tailored to suit the interests of teenagers. Management hypothesizes that there are no differences in the preferences for the four types of books. A sample of 1600 teenagers indicated the following preferences for the four book types. Publication Frequency of Preference Non-Fiction 515 Mystery 385 Humor 290 Short Stories 410 TOTAL 1600 Management needs your expertise to determine whether there are differences in preferences for the four types of books. Hint: This problem is the same format as the M&Ms example discussed in class. a. State the null and alternate hypotheses in plain English sentences (not equations). b. What is the calculated chi-squared value? Show your calculations. c. The tabled chi-squared value for 5% significance and three degrees of freedom is 7.81. Can you reject the null hypothesis? Why or why not?

Paper For Above instruction

Estimating the average number of hours professors work weekly is a common research goal in academic productivity studies. To achieve a reliable estimate with a specified confidence level and precision, one must determine the minimum sample size needed, considering the population variance. This paper addresses the calculation of required sample sizes under different conditions, examines a chi-squared test for independence regarding teenagers’ reading preferences, and explores a hypothesis test about consumer preferences for new frozen pizza products, as well as an analysis of the relationship between household income and car ownership.

Sample Size Calculation for Estimating Mean Hours Worked

The fundamental formula for determining the required sample size (n) when estimating a population mean is given by:

n = (Z^2 * σ^2) / E^2

where Z is the z-score corresponding to the desired confidence level, σ^2 is the population variance, and E is the margin of error (precision).

Given that the population variance (σ^2) is 90, and the desired confidence level is 95% with z=1.96, and the margin of error is ±2 hours, we can calculate the minimum sample size as follows:

n = (1.96^2 90) / 2^2 = (3.8416 90) / 4 = 345.744 / 4 ≈ 86.44

Therefore, a minimum sample size of 87 professors is required to meet these criteria.

If the precision is doubled to ±1 hour, then E=1. The sample size becomes:

n = (1.96^2 90) / 1^2 = (3.8416 90) / 1 = 345.744 / 1 ≈ 346

This demonstrates that increasing the accuracy (reducing the margin of error) substantially increases the required sample size.

Furthermore, increasing the confidence level to 99% with z=2.575 and returning the precision to ±2 hours yields:

n = (2.575^2 90) / 4 = (6.630625 90) / 4 = 596.756 / 4 ≈ 149

This indicates that higher confidence levels also result in larger sample size requirements to maintain the same precision.

Chi-Squared Test for Independence in Teenagers’ Reading Preferences

The second part of the analysis involves testing whether there are differences in teenagers’ preferences among four types of books: Non-Fiction, Mystery, Humor, and Short Stories.

The null hypothesis (H₀) is that there are no differences in preferences among the four books. The alternative hypothesis (H₁) is that at least one book type has a different preference distribution. To test this, a chi-squared test for independence is appropriate.

The observed frequencies are:

  • Non-Fiction: 515
  • Mystery: 385
  • Humor: 290
  • Short Stories: 410

The total sample size is 1600.

Expected frequencies under the null hypothesis are calculated based on the assumption that preferences are uniformly distributed:

Expected for each category = (row total) * (column total) / overall total. Since the null hypothesis posits no difference, the expected frequency for each would be based on the proportion assuming equal preference, which is 1/4 for each of the four categories:

Expected frequency for each = 1600 / 4 = 400.

The chi-squared statistic is calculated as:

χ² = Σ[(Observed - Expected)² / Expected]

Calculating each term:

  • Non-Fiction: (515 - 400)² / 400 = (115)² / 400 = 13225 / 400 = 33.0625
  • Mystery: (385 - 400)² / 400 = (-15)² / 400 = 225 / 400 = 0.5625
  • Humor: (290 - 400)² / 400 = (-110)² / 400 = 12100 / 400 = 30.25
  • Short Stories: (410 - 400)² / 400 = (10)² / 400 = 100 / 400 = 0.25

Total chi-squared value:

χ² = 33.0625 + 0.5625 + 30.25 + 0.25 = 64.125

Comparing this to the critical value of 7.81 at 3 degrees of freedom and 5% significance level, the calculated χ² is significantly higher. Therefore, we reject the null hypothesis, indicating that there are differences in teenagers' preferences for the four types of books.

Preference for a New Frozen Pizza

In the third case, we evaluate whether the proportion of consumers interested in a new pizza exceeds 13%.

Null hypothesis (H₀): The true proportion of consumers who prefer the new pizza is 13% or less.

Alternative hypothesis (H₁): The true proportion exceeds 13%.

Sample size n=1000, with 161 indicating preference. The sample proportion (p̂) is:

p̂ = 161 / 1000 = 0.161

The standard error (SE) of the proportion under the null hypothesis p₀=0.13 is:

SE = √[p₀(1 - p₀) / n] = √[0.13 * 0.87 / 1000] ≈ √[0.1131 / 1000] ≈ √0.0001131 ≈ 0.0106

The z-statistic is calculated as:

z = (p̂ - p₀) / SE = (0.161 - 0.13) / 0.0106 ≈ 0.031 / 0.0106 ≈ 2.92

At a 95% significance level for a one-tailed test, the critical z-value is 1.645. Since 2.92 > 1.645, we reject H₀ and conclude that more than 13% of consumers prefer the new pizza.

Relationship Between Income and Car Ownership

The final analysis investigates whether household income level is associated with possessing more than one car. The hypotheses are:

  • Null hypothesis (H₀): Income level is independent of whether a household has more than one car.
  • Alternative hypothesis (H₁): Income level is associated with having more than one car.

The observed frequencies are:

  • Low income, one car: 40
  • Low income, more than one car: 16
  • High income, one car: 15
  • High income, more than one car: 29

Total families: 100.

Calculating expected frequencies under independence involves:

  • Row totals:
  • Low income: 40 + 16 = 56
  • High income: 15 + 29 = 44
  • Column totals:
  • One car: 40 + 15 = 55
  • More than one car: 16 + 29 = 45
  • Expected frequency for each cell = (row total * column total) / overall total.

For example, for low income & one car:

Expected = (56 * 55) / 100 = 30.8

Similarly:

  • Low income & >1 car: (56 * 45) / 100 = 25.2
  • High income & one car: (44 * 55) / 100 = 24.2
  • High income & >1 car: (44 * 45) / 100 = 19.8

The chi-squared statistic is:

χ² = Σ[(Observed - Expected)² / Expected]

Calculations:

  • Low income & 1 car: (40 - 30.8)² / 30.8 ≈ (9.2)² / 30.8 ≈ 84.64 / 30.8 ≈ 2.75
  • Low income & >1 car: (16 - 25.2)² / 25.2 ≈ (-9.2)² / 25.2 ≈ 84.64 / 25.2 ≈ 3.36
  • High income & 1 car: (15 - 24.2)² / 24.2 ≈ (-9.2)² / 24.2 ≈ 84.64 / 24.2 ≈ 3.50
  • High income & >1 car: (29 - 19.8)² / 19.8 ≈ (9.2)² / 19.8 ≈ 84.64 / 19.8 ≈ 4.28

Total chi-squared value:

χ² ≈ 2.75 + 3.36 + 3.50 + 4.28 = 13.89

Comparing this to the critical value for 1 degree of freedom at 5% significance, which is 3.84, the calculated χ² exceeds this threshold. Therefore, we reject the null hypothesis, indicating a significant association between income level and possessing more than one car.

Conclusion

This analysis demonstrates the importance of sample size determination, hypothesis testing, and chi-squared analysis in understanding population parameters and relationships. Adequate sample sizes ensure reliable estimates; chi-squared tests help assess independence and relationships between categorical variables; and proportion tests are crucial for market research decisions. Recognizing the impacts of confidence levels, precision, and significance thresholds guides researchers in making informed conclusions in social sciences and business analytics.

References

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