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Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used.
Find the proportion in a t-distribution less than - 1.4 if the samples have sizes $n_{1}=30$ and $n_{2}=40$.
Enter the exact answer for the degrees of freedom and round your answer for the area to three decimal places.
degrees of freedom $=$
proportion $=$
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Answer
Final Answer: The degrees of freedom is \(\boxed{68}\) and the proportion in a t-distribution less than -1.4 is approximately \(\boxed{0.083}\).
Step 1 ：Given the sample sizes \(n_{1}=30\) and \(n_{2}=40\), we can calculate the degrees of freedom as \(df = n_{1} + n_{2} - 2\).
Step 2 ：Substituting the given values, we get \(df = 30 + 40 - 2 = 68\).
Step 3 ：We are asked to find the proportion in a t-distribution less than -1.4. This can be calculated using the cumulative distribution function (CDF) of the t-distribution with the calculated degrees of freedom.
Step 4 ：Using the CDF, we find that the proportion is approximately 0.08302968110666753.
Step 5 ：Rounding this to three decimal places, we get a proportion of 0.083.
Step 6 ：Final Answer: The degrees of freedom is \(\boxed{68}\) and the proportion in a t-distribution less than -1.4 is approximately \(\boxed{0.083}\).
