Question

two samples are taken with the following sample means, standard deviations, and sample sizes.
$\bar{x}_1 = 34$
$s_1 = 4$
$n_1 = 72$
$\bar{x}_2 = 39$
$s_2 = 2$
$n_2 = 47$
estimate the difference in population means using a 87% confidence level. use a calculator.
round answers to 1 decimal place.

Explanation:

Step1: Determine the degrees of freedom

Since the sample sizes $n_1 = 72$ and $n_2=47$ are large (usually $n>30$ is considered large - sample case), we can approximate the degrees - of - freedom using the larger sample size. Here, we can use a normal distribution approximation. The confidence level is $87\%$, so the significance level $\alpha=1 - 0.87 = 0.13$. Then $\alpha/2=0.065$.

Step2: Find the z - value

Looking up the $z$ - value in the standard normal distribution table, $z_{\alpha/2}=z_{0.065}$. The $z$ - value corresponding to a right - tail area of $0.065$ is approximately $z = 1.51$.

Step3: Calculate the difference in sample means

The difference in sample means is $\bar{x}_1-\bar{x}_2=34 - 39=- 5$.

Step4: Calculate the standard error

The formula for the standard error of the difference in means for two independent samples is $SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}$. Substitute $s_1 = 4$, $n_1 = 72$, $s_2 = 2$, and $n_2 = 47$ into the formula:

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Step5: Calculate the margin of error

The margin of error $E = z_{\alpha/2}\times SE$. Substitute $z_{\alpha/2}=1.51$ and $SE\approx0.554$ into the formula: $E = 1.51\times0.554\approx0.836$.

Step6: Calculate the confidence interval

The confidence interval for the difference in population means $\mu_1-\mu_2$ is $(\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E$. Substitute $\bar{x}_1-\bar{x}_2=-5$ and $E\approx0.836$:

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Answer:

$(-5.8,-4.2)$