Mathematics, 05.05.2020 04:01 kellersweeney

The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing that is said to emulate the legendary champion, Byron Nelson. Ten randomly selected balls of two brands are tested and the overall distance measured. Assume that the variances are equal. The data follow: Brand 1: 276 279 287 271 283 271 279 275 263 267 Brand 2: 270 238 260 265 273 281 271 270 263 268 (a) Is there evidence to show that there is a difference in the mean overall distance of brands? Use α=0.05. Choose your answer in accordance to the item a) of the question statement (b) Calculate a 95% two-sided confidence interval on the difference in mean overall distance. Round your answer to one decimal place (e. g. 98.7). Enter your answer; confidence interval, lower bound ≤μ1-μ2≤ Enter your answer; confidence interval, upper bound

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Answer from: lizbethmillanvazquez

a) There is enough evidence to support the claim that there is a difference in the mean overall distance of brands.

b) The 95% confidence interval for the mean is (0.8, 17.6).

$0.8\leq\mu_1-\mu_2\leq17.6$

Step-by-step explanation:

The data follow:

Brand 1: 276 279 287 271 283 271 279 275 263 267

Brand 2: 270 238 260 265 273 281 271 270 263 268

Mean and STD for Brand 1

$M_1=\dfrac{1}{10}\sum_{i=1}^{10}(276+279+287+271+283+271+279+275+263+267)\\\\\\ M_1=\dfrac{2751}{10}=275.1$

$s_1=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\s_1=\sqrt{\dfrac{1}{9}\cdot [(276-(275.1))^2+...+(267-(275.1))^2]}\\\\\\$

$s_1=\sqrt{\dfrac{1}{9}\cdot [(0.81)+(15.21)+...+(65.61)]}=\sqrt{\dfrac{480.9}{9}}=\sqrt{53.43}\\\\\\s_1=7.31$

Mean and STD for Brand 2

$M_2=\dfrac{1}{10}\sum_{i=1}^{10}(270+238+260+265+273+281+271+270+263+268)\\\\\\ M_2=\dfrac{2659}{10}=265.9$

$s_2=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{10}(x_i-M)^2}\\\\\\s_2=\sqrt{\dfrac{1}{9}\cdot [(270-(265.9))^2+(238-(265.9))^2+...+(268-(265.9))^2]}\\\\\\$

$s_2=\sqrt{\dfrac{1}{9}\cdot [(16.81)+(778.41)+...+(4.41)]}=\sqrt{\dfrac{1164.9}{9}}=\sqrt{129.43}\\\\\\s_2=11.38$

This is a hypothesis test for the difference between populations means.

The claim is that there is a difference in the mean overall distance of brands.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2 0$

The significance level is 0.05.

The sample 1, of size n1=10 has a mean of 275.1 and a standard deviation of 7.31.

The sample 1, of size n1=10 has a mean of 265.9 and a standard deviation of 11.38.

The difference between sample means is Md=9.2.

M_d=M_1-M_2=275.1-265.9=9.2

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{7.31^2}{10}+\dfrac{11.38^2}{10}}\\\\\\s_{M_d}=\sqrt{5.344+12.95}=\sqrt{18.294}=4.277$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{9.2-0}{4.277}=\dfrac{9.2}{4.277}=2.151$

The degrees of freedom for this test are:

df=n_1+n_2-1=10+10-2=18

This test is a right-tailed test, with 18 degrees of freedom and t=2.151, so the P-value for this test is calculated as (using a t-table):

P-value=P(t2.151)=0.023

As the P-value (0.023) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that there is a difference in the mean overall distance of brands.

b) We have to calculate a 95% confidence interval for the difference between means, with a T-model.

The difference between sample means is Md=9.2.

The estimated standard error of the difference is s_Md=4.277.

The critical t-value for a 95% confidence interval is t=1.96.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.96 \cdot 4.277=8.383$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 9.2-8.383=0.817\\\\UL=M_d+t \cdot s_{M_d} = 9.2+8.383=17.583$

The 95% confidence interval for the mean is (0.8, 17.6).

$0.8\leq\mu_1-\mu_2\leq17.6$

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