Mathematics, 31.07.2020 09:01 sammimoose6490

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. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.01 significance level for both parts. Diet Regular
μ μ1 μ2
n 20 20
x 0.78062lb 0.81645 lb
s 0.00444 lb 0.00745 lb
A. Test the claim that the contents of cans of diet soda have weights with a mean that is less than the mean for the regular soda.
What are the null and alternative and hypotheses?
B. What is the test statistic? (Round to two decimal places as needed.)
C. What is the P-value? (Round to three decimal places as needed.)
State the conclusion for the test.
A. 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.
B. Fail to 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.
C. Fail to reject the null hypothesis. There is not 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.
D. Reject the null hypothesis. There is not 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.
b. Construct a confidence interval appropriate for the hypothesis test in part (a).
_lb < u1 - u2 < _lb (Round to three decimal places as needed.)
Does the confidence interval support the conclusion found with the hypothesis test?
(No/Yes) because the confidence interval contains (zero/only positives values/ only negative values)

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Answer from: monet83

(A) Null Hypothesis, H_0 : $\mu_1 \geq \mu_2$

Alternate Hypothesis, H_A : $\mu_1$

(B) The value of t-test statistics is -18.48.

(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 $\mu_1$ = mean weight of contents of cans of diet soda.

$\mu_2$ = mean weight of contents of cans of regular soda.

So, Null Hypothesis, H_0 : $\mu_1 \geq \mu_2$ {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, H_A : $\mu_1$ {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. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ t__n_1_+_n_2_-_2

where, $\bar X_1$ = sample mean weight of cans of diet soda = 0.78062 lb

$\bar X_2$ = sample mean weight of cans of regular soda = 0.81645 lb

s_1 = sample standard deviation of cans of diet soda = 0.00444 lb

s_2 = sample standard deviation of cans of regular soda = 0.00745 lb

n_1 = sample of cans of diet soda = 20

n_2 = sample of cans of diet soda = 20

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(20-1)\times 0.00444^{2}+ (20-1)\times 0.00745^{2}}{20+20-2} }$ = 0.00613

So, the test statistics = $\frac{(0.78062-0.81645)-(0)}{0.00613 \times \sqrt{\frac{1}{20}+\frac{1}{20} } }$ ~ t_3_8

= -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( t_3_8 < -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.

Answer from: Quest

on your paper it doesn't say the answer but the the answer is altitude

step-by-step explanation:

What is the name of the segment inside the large triangle?

Answer from: Quest

Then why is the question on here if you know the answer

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