Research Conducted By The University Of

Research Conducted By The University Of

Discussion-2_STAT-101 A research conducted by the University of Michigan claimed that there are more female drivers in the USA than male drivers. A researcher decides to test this claim on his state. In his simple random sample of 400 observations, he noticed that 210 of 400 were women. At α = 0.05, is there enough evidence to support the claim? Hint: Null and alternative hypotheses are ; (Use the critical value method where the critical value of z for 0.05 significance level is 1.645 )

Paper For Above instruction

Introduction

The issue of gender disparities in driving populations has been a subject of ongoing research, with some studies suggesting that women may outnumber men among drivers in the United States. To evaluate this claim at the state level, a researcher conducted a statistical hypothesis test based on a sample of 400 drivers, observing that 210 of these were women. The central question is whether the data provides sufficient evidence to support the hypothesis that female drivers are more numerous than male drivers at a significance level of 0.05. This paper discusses the appropriate null and alternative hypotheses, the statistical methodology, calculations, and the conclusion derived from the hypothesis test.

Null and Alternative Hypotheses

The research claim focuses on the proportion of female drivers. Let p be the true proportion of female drivers in the population. The hypotheses are formulated as:

• Null hypothesis (H₀): p = 0.5 (The proportion of female drivers is equal to 50%)
• Alternative hypothesis (H₁): p > 0.5 (More than 50% of drivers are female)

This formulation aligns with the claim that females outnumber males among drivers. Conducting a one-tailed test at α = 0.05 allows us to determine whether the observed data is sufficiently extreme to reject H₀ in favor of H₁.

Methodology

The test involves comparing the observed sample proportion with the hypothesized population proportion. The sample proportion (\( \hat{p} \)) is calculated as:

\( \hat{p} = \frac{210}{400} = 0.525 \)

Using the standard normal (z) test for proportions, the test statistic (z) is computed as:

\( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \)

where \( p_0 = 0.5 \) and \( n = 400 \).

Calculation of the z-value:

\( z = \frac{0.525 - 0.5}{\sqrt{\frac{0.5 \times 0.5}{400}}} = \frac{0.025}{\sqrt{\frac{0.25}{400}}} = \frac{0.025}{\sqrt{0.000625}} = \frac{0.025}{0.025} = 1.0 \)

Critical Value and Decision

The critical z-value at a significance level of 0.05 for a one-tailed test is 1.645, as specified. Since our calculated z-value of 1.0 is less than 1.645, we do not reject the null hypothesis. This indicates that, based on the sample data, there is insufficient evidence to conclude that the proportion of female drivers exceeds 50%.

Conclusion

In conclusion, the statistical analysis suggests that the observed difference in sample proportion (52.5%) is not statistically significant at the 5% significance level. Therefore, there is no enough evidence to support the claim that female drivers outnumber male drivers in the state's population. Further research with larger samples or different methodologies could provide more insight into the gender distribution among drivers.

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