Respuesta :
Answer:
Yes. There is enough evidence to support the claim that there are fewer errors on average when the yellow ball is used.
Step-by-step explanation:
The question is incomplete:
The sample data is:
Yellow 5 2 6 7 2 5 3 8 4 9
White 7 6 8 5 9 11 8 3 6 10
This is a hypothesis test for the difference between populations means.
The claim is that there are fewer errors on average when the yellow ball is used.
Then, the null and alternative hypothesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0[/tex]
The significance level is α=0.05.
The sample 1 (yellow ball errors), of size n1=10 has a mean of 5.1 and a standard deviation of 2.42.
The sample 2 (white balls errors), of size n2=10 has a mean of 7.3 and a standard deviation of 2.41.
The difference between sample means is Md=-2.2.
[tex]M_d=M_1-M_2=5.1-7.3=-2.2[/tex]
The estimated standard error of the difference between means is computed using the formula:
[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{2.42^2+2.41^2}{10}}\\\\\\s_{M_d}=\sqrt{\dfrac{11.665}{10}}=\sqrt{1.166}=1.08[/tex]
Then, we can calculate the t-statistic as:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-2.2-0}{1.08}=\dfrac{-2.2}{1.08}=-2.037[/tex]
The degrees of freedom for this test are:
[tex]df=n_1+n_2-1=10+10-2=18[/tex]
This test is a left-tailed test, with 18 degrees of freedom and t=-2.037, so the P-value for this test is calculated as (using a t-table):
[tex]P-value=P(t<-2.037)=0.028[/tex]
As the P-value (0.028) 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 are fewer errors on average when the yellow ball is used.
Yes, there are fewer errors on average when the yellow ball is used and this can be determined by using the given data.
The Hypothesis test is carried out in which null and alternate hypothesis is given below:
[tex]\rm H_0 : \mu_1-\mu_2=0[/tex]
[tex]\rm H_a : \mu_1-\mu_2<0[/tex]
Now, determine the sample mean difference.
[tex]\rm M_d = M_1-M_2 = 5.1-7.3 = -2.2[/tex]
Now, determine the estimated standard error using the below formula:
[tex]\rm s =\sqrt{\dfrac{\sigma^2_1+\sigma^2_2}{n}}[/tex]
[tex]\rm s =\sqrt{\dfrac{(2.42)^2+(2.41)^2}{10}}[/tex]
s = 1.08
So, the t-statistics can be calculated as:
[tex]\rm t = \dfrac{M_d-(\mu_1-\mu_2)}{s}[/tex]
[tex]\rm t = \dfrac{-2.2-0}{1.08}=-2.037[/tex]
Now, determine the degree of freedom.
[tex]\rm df = n_1+n_2-1[/tex]
df = 10 + 10 - 2
df = 18
Now, for this test, the p-value is 0.028 which is less than the significance level. Therefore, the null hypothesis is rejected.
For more information, refer to the link given below:
https://brainly.com/question/4454077
