Question

z-Tests According to the CDC the average height of 15-year-old boys is 67 inches, with a standard deviation of 3.19 inches. A high school teacher believes the male students in his classes are especially tall this year. The teacher averages the heights of a random sample of the male students in his classes throughout the day (N 57) and find an average of 68.1 inches. 1. Use a two-tailed hypothesis (z) test with alpha0.01 to determine if there is any difference in average height of this teacher's students and all 15-year-old boys in the CDC database. State the following in your answer: A) Your null and alternative hypotheses, B) z-crit, and z-observed, C) Final statistical decision. Interpret your decision in a complete sentence. 2. Calculate a 99% confidence interval. State a decision in regard to the null hypothesis in #1 based on this interval. Interpret your 99% confidence interval. 3. Calculate a 95%.confidence intervalinstead. How is it different than the 99% confidence interval you calculated in #2? 4. Calculate effect size using Cohen's d. Classify the effect size as small, medium, or large AND interpret your result.

Answer 1

1)

A) The null and alternate hypotheses are

$H_0:\mu _1-\mu _2=0\\ H_1:\mu _1-\mu _2\neq 0$

B) Here $\alpha =0.01$ . So the critical values assuming normal distribution is

$\pm z_{1-\alpha /2}={\color{Blue} \pm 2.576}$

C) The sample means are $\overline{x_1}=67,\overline{x_2}=68.1$ The sample standard deviations are

$s_1=s_2=3.19$ and $n=57$ .

The estimated mean difference is $\overline{x_1}-\overline{x_2}=67-68.1=-1.1$

The sample standard deviation of the differences is   $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=\sqrt{\frac{3.19^2}{57}+\frac{3.19^2}{57}}=0.5975$

The test statistic/ observed z is

$\left (\overline{x_1}-\overline{x_2} \right )/\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}=-1.1/0.5975=-1.841$

Since   $\left | \pm 2.576 \right |<\left | -1.841 \right |$ , the test statistic does not lie in the critical region, and we accept the null hypothesis.There is no differenec in heights at 99% confidence level.

2) The $\left ( 1-\alpha \right )100\%$ CI for difference in means is mean is

$\overline{x_1}-\overline{x_2} \pm z_{1-\alpha /2}\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\$

The 99% confidence interval for the mean difference is ( $\alpha =0.01$ )

$-1.1\pm 2.5758 \left (0.5975\right )\\ {\color{Blue} \left (-2.6390578, 0.4390578 \right )}$

Since the CI includes 0, $0\in \left (-2.6390578, 0.4390578 \right )$ , we accept te null hypothesis. The difference in mean lies in the interval   $\left (-2.6391, 0.4391\right )$ at 99% confidence.

3)The 95% confidence interval for the mean difference is ( $\alpha =0.05$ )

$-1.1\pm1.96\left (0.5975\right )\\ {\color{Blue} \left ( -2.2711 , 0.0711 \right )}$

The confidence interval has become narrower than that of 99% CI.

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