Respuesta :
Answer:
(A) Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \geq \mu_2[/tex]
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1<\mu_2[/tex]
(B) The value of t-test statistics is -18.48.
(C) The P-value is Less than 0.005%.
(D) Reject the null hypothesis. There is sufficient evidence to support the claim that the cans of diet soda have mean weights that are lower than the mean weight for the regular soda.
Step-by-step explanation:
We are given that the Data on the weights (lb) of the contents of cans of diet soda versus the contents of cans of the regular version of the soda is summarized to the right;
Diet Regular
μ μ1 μ2
n 20 20
x 0.78062lb 0.81645 lb
s 0.00444 lb 0.00745 lb
Let [tex]\mu_1[/tex] = mean weight of contents of cans of diet soda.
[tex]\mu_2[/tex] = mean weight of contents of cans of regular soda.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu_1 \geq \mu_2[/tex] {means that the contents of cans of diet soda have weights with a mean that is more than or equal to the mean for the regular soda}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu_1<\mu_2[/tex] {means that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda}
The test statistics that will be used here is Two-sample t-test statistics because we don't know about population standard deviations;
T.S. = [tex]\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ~ [tex]t__n_1_+_n_2_-_2[/tex]
where, [tex]\bar X_1[/tex] = sample mean weight of cans of diet soda = 0.78062 lb
[tex]\bar X_2[/tex] = sample mean weight of cans of regular soda = 0.81645 lb
[tex]s_1[/tex] = sample standard deviation of cans of diet soda = 0.00444 lb
[tex]s_2[/tex] = sample standard deviation of cans of regular soda = 0.00745 lb
[tex]n_1[/tex] = sample of cans of diet soda = 20
[tex]n_2[/tex] = sample of cans of diet soda = 20
Also, [tex]s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(20-1)\times 0.00444^{2}+ (20-1)\times 0.00745^{2}}{20+20-2} }[/tex] = 0.00613
So, the test statistics = [tex]\frac{(0.78062-0.81645)-(0)}{0.00613 \times \sqrt{\frac{1}{20}+\frac{1}{20} } }[/tex] ~ [tex]t_3_8[/tex]
= -18.48
The value of t-test statistics is -18.48.
Also, the P-value of the test statistics is given by;
P-value = P( [tex]t_3_8[/tex] < -18.48) = Less than 0.005%
Now, at a 0.01 level of significance, the t table gives a critical value of -2.429 at 38 degrees of freedom for the left-tailed test.
Since the value of our test statistics is less than the critical value of t as -18.48 < -2.429, so we have sufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda.