Question

1. consider the following sample data of 3,310 patients and their cholesterol levels. men women sample size 1544 1766 mean 192.4 207.1 standard deviation 35.2 36.7 a) (5 points) calculate the 95% confidence interval for the difference in population cholesterol levels between men and women. you can assume that the population variance for men and women cholesterol levels is equal. include a written interpretation of your confidence interval.

Answer 1

The value of t = 1.960681

We have to find the value of a t.

For men,

Sample size ($n_{1}$) = 1544

Sample mean ($x_{1}$) = 192.4

Standard deviation ($s_{1}$) = 35.2

For women,

Sample size ($n_{2}$) = 1766

Sample mean ($x_{2}$) = 207.1

Standard deviation ($s_{2}$) $= 36.7$

Confidential level (c) = 95% = 0.95

Level of significance ($\alpha$) = 0.05

To find the t critical value corresponding to the probability 0.05 and

Degrees of freedom is (df) = $n_{1}+n_{2}-2=1544+1766-2=3308$

We use below excel syntax.

=TINV(probability, degrees of freedom)

Please type in any empty cell of excel == TINV(0.05, 3308) then hit entre button.

The required z value is t = 1.960681.

(a).   The required 95% confidence interval for the difference in population cholesterol levels between men and women is given by,

$\text { C.I }=\left(\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_2\right) \pm \mathrm{t} \times \mathrm{s}_{\mathrm{p}} \sqrt{\left(\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}\right)}$

Where,

$\begin{aligned}\mathrm{s}_{\mathrm{p}} & =\sqrt{\frac{\left(\mathrm{n}_1-1\right) \times \mathrm{s}_1^2+\left(\mathrm{n}_2-1\right) \times \mathrm{s}_2^2}{\mathrm{n}_1+\mathrm{n}_2-2}} \\& =\sqrt{\frac{(1,544-1) \times 35.2^2+(1,766-1) \times 36.7^2}{1,544+1,766-2}} \\& =\sqrt{1,296.584} \\& =36.00811\end{aligned}$

Therefore,

$\begin{aligned}\text { C.I } & =\left(\overline{\mathrm{x}}_1-\overline{\mathrm{x}}_2\right) \pm \mathrm{t} \times \mathrm{s}_{\mathrm{p}} \sqrt{\left(\frac{1}{\mathrm{n}_1}+\frac{1}{\mathrm{n}_2}\right)} \\& =(192.4-207.1) \pm 1.960681 \times 36.00811 \sqrt{\left(\frac{1}{1,544}+\frac{1}{1,766}\right)} \\& =(-14.7) \pm 1.960681 \times 36.00811 \times 0.034841 \\& =-14.7 \pm 2.459814 \\& =(-17.16,-12.24)\end{aligned}$

The required 95% confidence interval for the difference in population cholesterol levels between men and women is (-17.16, -12.24)

(b). We are given,

Null hypothesis $\left(H_0\right): \mu_M-\mu_{\mathrm{W}}=0$

According to the answer of part a), If the confidence interval does not include the null value (zero), so we reject the null hypothesis.

We conclude that there is a statistically significant difference between the groups.

Conclusion

Reject$H_{0}$. There is significance difference between the men and women have different cholesterol levels.

For more questions on null hypothesis

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