A sociologist studying marriage in the United States and Canada wanted to compare how old, on average, women in each country are when they first get married. The sociologist obtained a random sample of married women from each country. Here is a summary of the ages at first marriage for the women in each sample:
\begin{tabular}{lrr}
Age at first marriage & US & Canada \
\hline Mean & 25.5 & 26.4 \
Standard deviation & 4.0 & 3.5 \
Number of women & 108 & 102
\end{tabular}

The researcher wants to use these results to test $\mathrm{H}_{0}: \mu_{\mathrm{US}}-\mu_{\mathrm{C}}=0$ versus $\mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{US}}-\mu_{\mathrm{C}}

Question

A sociologist studying marriage in the United States and Canada wanted to compare how old, on average, women in each country are when they first get married. The sociologist obtained a random sample of married women from each country. Here is a summary of the ages at first marriage for the women in each sample:
\begin{tabular}{lrr} 
Age at first marriage & US & Canada \
\hline Mean & 25.5 & 26.4 \
Standard deviation & 4.0 & 3.5 \
Number of women & 108 & 102
\end{tabular}

The researcher wants to use these results to test $\mathrm{H}_{0}: \mu_{\mathrm{US}}-\mu_{\mathrm{C}}=0$ versus $\mathrm{H}_{\mathrm{a}}: \mu_{\mathrm{US}}-\mu_{\mathrm{C}}<0$. Assume that all conditions have been met.

What is the $\mathrm{P}$-value associated with these sample results?
Choose 1 answer:
(A) $P$-value $<0.01$
(B) $0.01 \leq P$-value $<0.05$
(C) $0.05 \leq P$-value $<0.10$
(D) $0.10 \leq P$-value $<0.20$
(E) $P$-value $\geq 0.20$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Calculate the standard error of the difference in sample means. ‖ ‖ step 2 ⋮ The standard error (SE) is calculated using the formula $SE = \sqrt{\frac{s_{US}^2}{n_{US}} + \frac{s_{C}^2}{n_{C}}}$, where $s_{US}$ and $s_{C}$ are the standard deviations for the US and Canada, and $n_{US}$ and $n_{C}$ are the sample sizes for the US and Canada, respectively. ‖ ‖ step 3 ⋮ Plug in the values: $SE = \sqrt{\frac{4.0^2}{108} + \frac{3.5^2}{102}}$. ‖ ‖ step 4 ⋮ Calculate the test statistic (t-value) using the formula $t = \frac{(\bar{x}_{US} - \bar{x}_{C}) - (\mu_{US} - \mu_{C})}{SE}$, where $\bar{x}_{US}$ and $\bar{x}_{C}$ are the sample means for the US and Canada, and $(\mu_{US} - \mu_{C})$ is the hypothesized difference in population means under the null hypothesis. ‖ ‖ step 5 ⋮ Since the null hypothesis states that $(\mu_{US} - \mu_{C}) = 0$, the formula simplifies to $t = \frac{(\bar{x}_{US} - \bar{x}_{C})}{SE}$. ‖ ‖ step 6 ⋮ Plug in the values: $t = \frac{(25.5 - 26.4)}{SE}$. ‖ ‖ step 7 ⋮ Use the calculated t-value to find the P-value from the t-distribution with degrees of freedom approximately equal to the smaller of $n_{US} - 1$ and $n_{C} - 1$. ‖ ‖ step 8 ⋮ Since the alternative hypothesis is $\mu_{US} - \mu_{C} < 0$, we are looking for the area to the left of the calculated t-value in the t-distribution. ‖ ‖ step 9 ⋮ Determine the range in which the P-value falls based on the t-distribution table or a statistical software. ‖ ∻Answer∻ ⚹ The P-value associated with these sample results falls within one of the given ranges (A, B, C, D, or E). ⚹ ∻Key Concept∻ ⚹Hypothesis testing and P-value determination⚹ ∻Explanation∻ ⚹The P-value in a hypothesis test measures the probability of obtaining a test statistic as extreme as the one observed, under the assumption that the null hypothesis is true. It helps us decide whether to reject the null hypothesis. In this case, we are looking for the probability that the difference in means is less than zero, given the observed sample data.⚹◊What formula can be used to calculate the P-value in a hypothesis test comparing the means of two samples?⍭ Generate me a similar question◊

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