Time Married Men With Children Spend On Child Care Average
The Time Married Men With Children Spend On Child Care Averages 64 Ho
The time married men with children spend on child care averages 6.4 hours per week (Time, March 12, 2012). You belong to a professional group on family practices that would like to do its own study to determine if the time married men in your area spend on child care per week differs from the reported mean of 6.4 hours per week. A sample of 40 married couples will be used with the data collected showing the hours per week the husband spends on child care. Sample mean is 7, sample stdv. is 2.46.
a. What are the hypotheses if your group would like to determine if the population mean number of hours married men are spending in child care differs from the mean reported by Time in your area?
b. What is the sample mean and the p-value?
c. Select your own level of significance. What is your conclusion?
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Paper For Above instruction
Introduction
Understanding how much time married men spend on child care is essential for assessing family dynamics and informing policy decisions. Recent studies, such as the report by Time (2012), indicate that married men with children spend an average of 6.4 hours weekly on child care. However, regional variations and changing societal roles necessitate localized investigations. This study aims to determine whether the actual average time spent by married men in our area differs significantly from the reported national mean. Using a sample of 40 married couples, we compare the observed data against the known or assumed parameters, applying statistical hypothesis testing techniques suited to the available data.
Formulating the Hypotheses
The primary objective is to assess whether the population mean time married men spend on child care in our region differs from the reported national average of 6.4 hours. The hypotheses are constructed as follows:
- Null hypothesis (H0): The population mean time spent on child care by married men in our area equals 6.4 hours. Mathematically, H0: μ = 6.4.
- Alternative hypothesis (H1): The population mean time differs from 6.4 hours. Mathematically, H1: μ ≠ 6.4.
This two-tailed hypothesis test evaluates whether the observed sample mean provides sufficient evidence to conclude a significant difference exists.
Data Summary
The sample comprises 40 married men, with the sample mean (\(\bar{x}\)) calculated as 7 hours and the sample standard deviation (s) as 2.46 hours. The sample size (n) is 40. The observed mean exceeds the reported mean, indicating a potential difference, but statistical testing is required to determine significance.
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Part 1: Assuming Known Population Standard Deviation
When the population standard deviation (\(\sigma\)) is known, the Z-test for the mean applies. Given \(\sigma = 2.5\) hours and the significance level (\(\alpha\)) set at 0.05, the test proceeds as follows:
Test Statistic Calculation
\[
Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} = \frac{7 - 6.4}{2.5 / \sqrt{40}} \approx \frac{0.6}{2.5 / 6.3246} \approx \frac{0.6}{0.395} \approx 1.52
\]
Critical Value
Using the standard normal distribution table, the critical z-value for a two-tailed test at \(\alpha = 0.05\) is approximately ±1.96.
Decision Rule
Because the calculated z-value of 1.52 falls within the range (-1.96, 1.96), we do not reject H0 at the 0.05 significance level.
P-Value Calculation
Using the standard normal distribution, the p-value corresponds to twice the area beyond |1.52|:
\[
p = 2 \times P(Z > 1.52) \approx 2 \times 0.064 = 0.128
\]
Since \(p = 0.128 > 0.05\), there is insufficient evidence to reject the null hypothesis.
Conclusion
Based on the Z-test, we do not have enough statistical evidence at the 5% significance level to conclude that the average time married men in our area differs from the national average of 6.4 hours.
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Part 2: Assuming Unknown Population Standard Deviation
When \(\sigma\) is unknown, the t-test for the mean applies, accounting for additional uncertainty with sample standard deviation s.
Test Statistic Calculation
\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{7 - 6.4}{2.46 / \sqrt{40}} \approx \frac{0.6}{2.46 / 6.3246} \approx \frac{0.6}{0.389} \approx 1.54
\]
Degrees of Freedom
\(df = n - 1 = 39\)
Critical t-Value
At \(\alpha = 0.05\), the two-tailed critical t-value for df=39 is approximately ±2.022.
Decision Rule
Since 1.54
P-Value Calculation
Using statistical software or t-distribution tables:
\[
p = 2 \times P(T_{39} > 1.54) \approx 2 \times 0.064 \approx 0.128
\]
Again, p-value exceeds the significance level, indicating insufficient evidence to reject H0.
Business Conclusion
Both tests—assuming known and unknown population standard deviation—suggest that there is no statistically significant difference in the mean time spent on child care by married men in our region compared to the national average of 6.4 hours, at the 0.05 significance level. This indicates parity with broader findings and supports the notion that, regionally, men’s involvement in child care remains consistent with national reports.
Discussion
The consistency across the two scenarios underscores the robustness of the conclusion. Variability in individual circumstances and socio-economic factors may influence the mean, but this data does not provide compelling evidence for a regional deviation. Future studies could benefit from larger samples or stratification by demographic variables to uncover nuanced patterns.
Conclusion
The hypothesis testing indicates that, statistically, the average number of hours married men spend on child care in our area does not differ significantly from the reported national mean of 6.4 hours. These findings suggest homogeneity in paternal involvement across contexts, with implications for family policy, social interventions, and future research aimed at understanding gender roles within the family domain.
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References
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