Using the t-distribution, it is found that the 99% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model is (-25.4, 40.4).

We will find the standard deviation for each sample, hence, the t-distribution will be used.

For each sample, the mean, standard deviation and sample sizes are given by:

$\mu_1 = 259.6, s_1 = 18.72, n_1 = 10$

$\mu_2 = 252.1, s_2 = 30.95, n_2 = 10$

The standard error for each sample is given by:

$s_{e1} = \frac{18.72}{\sqrt{10}} = 5.92$

$s_{e2} = \frac{30.95}{\sqrt{10}} = 9.79$

The distribution of the differences has mean and standard error given by:

$\overline{x} = \mu_1 - \mu_2 = 259.6 - 252.1 = 7.5$

$s = \sqrt{s_{e1}^2 + s_{e2}^2} = \sqrt{5.92^2 + 9.79^2} = 11.44$

The interval is:

$\overline{x} \pm ts$

The critical value for a 99% two-tailed confidence interval, with 10 + 10 - 2 = 18 df, is t = 2.8784.

Then:

$\overline{x} - ts = 7.5 - 2.8784(11.44) = -25.4$

$\overline{x} + ts = 7.5 + 2.8784(11.44) = 40.4$

The 99% confidence interval for the true difference between the mean distance hit with the new model and the mean distance hit with the older model is (-25.4, 40.4).

To learn more about the use of the t-distribution to build a confidence interval, you can check brainly.com/question/25675821

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2 years ago

Whats 9+10 you answer right you get 20 points jk only 10

velikii [3]

Answer:19/21

Step-by-step explanation:

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3 years ago

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