Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month. A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces. Breyers would like to set LaTeX: \alpha = 0.025 α = 0.025 for the hypothesis test. It is known that LaTeX: z_{\alpha}=1.96 z α = 1.96 and LaTeX: t_{\alpha}=2.06 t α = 2.06 for the df = 24. Also, it is established that the ice cream consumption follows the normal distribution in the population. The conclusion for this hypothesis test would be

Answers

Answer 1

Answer:

Answer:

The conclusion for this hypothesis test would be that the average American consumes less than or equal to 17 ounces of ice cream per month.

Step-by-step explanation:

We are given that Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month.

A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.

Let $\mu$ = average ounces of ice cream consumed by American per month

So, Null Hypothesis, $H_0$ : $\mu \leq$ 17 ounces {means that the average American consumes less than or equal to 17 ounces of ice cream per month}

Alternate Hypothesis, $H_A$ : $\mu$ > 17 ounces {means that the average American consumes more than 17 ounces of ice cream per month}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $(\bar X-\mu)/((s)/(√(n) ) )$ ~ $t_n_-_1$

where, $\bar X$ = sample average = 19 ounces

s = sample standard deviation = 5 ounces

n = sample of Americans = 25

So, test statistics = $(19-17)/((5)/(√(25) ) )$ ~ $t_2_4$

= 2

The value of the test statistics is 2.

Now at 0.025 significance level, the t table gives critical value of 2.06 at 24 degree of freedom for right-tailed test. Since our test statistics is less than the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the the average American consumes less than or equal to 17 ounces of ice cream per month.

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Answers

Answer:

Sin A = 22/33.3 ≈ 0.7

Step-by-step explanation:

Sin A = opp/hyp

Opp = 22

Adj = 25

Apply pythagorean theorem to find the hypotenuse length.

Thus:

Hyp = √(25² + 22²) ≈ 33.3

Therefore,

Sin A = 22/33.3 ≈ 0.7 (nearest tenth)

A university was interested in examining the overall effectiveness of its online statistics course, along with the effectiveness of particular aspects of the course. First, the university wanted to see whether the online course was better than a standard course. Second, the university wanted to know whether students learned best using Excel, using RStudio, or using no statistical package at all. The university randomly selected a group of 30 students and administered one of the different variants of the course (i.e., traditional or online, coupled with one of the software options) to each student. The success of each variant was measured by the students' average improvement between a pre-test and a post-test. How many treatment groups are there in this study? a. 3
b. 4
c. 5
d. 6
e. 7

Answers

Answer:

Option D is correct - There are 6 treatment groups in this study.

Step-by-step explanation:

The number of treatment groups is equal to the number of possible combinations of the values of the factors.

In the question given, we have two factors: type of instruction (traditional/online) and software (Excel/Minitab/none).

Since there are 2 values for 'type of instruction' and 3 values for 'software'. Hence the number of treatment groups = 2*3 = 6.

Answer:

The answer is D: 5

Step-by-step explanation:

-Online course,

-Program to use,

- Number of students,

-Administered variant,

-Measurement of the average student.

One basketball team played 30 games throughout their entire season. If this team won 80% of those games, how many games did they win? Enter a numerical answer only.

Answers

the team won 24 games out of 30

Find three consecutive numbers whose sum is 612

Answers

Answer: 203, 204, 205

Step-by-step explanation:

Consecutive numbers are numbers that are right next to each other. An example would be 1,2,3 or 67,68,69. Since this problem is asking for three consecutive numbers where the sum is equal to 612, we need to find the numbers that will be greater than 200.

Since we know that the sum is 612, we know that the numbers will be in the 2 hundreds. All we need to do is find 3 numbers that equal to 12.

3+4+5=12

Now that we have the three numbers, let's add 200.

203+204+205=612

Therefore, the three numbers are 203, 204, 205.

Answer: 203, 204, 205

Step-by-step explanation:

Let's say the number is x.

So the three consecutive numbers are: x, x + 1, and x+ 2

3x + 3 = 612

3x = 609

x = 203

Plug in x.

203, 204, and 205

A restaurant has a capacity of 120 patrons. If the restaurant is full, how many patrons are at the restaurant?

Answers

Answer:120?

Step-by-step explanation:

Use the binomial theorem to expand the expression :(3x + y)^5 and simplify.
(b) find the middle term in the expansion of
(1/x+√x)^4 and simplify your unswer.
(c) determine the coefficient of x^11 in the expansion of (x^2 +1/x)^10, simplify your answer.

Answers

Answer:

a) $(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5$ .

b) The middle term in the expansion is $(6)/(x)$ .

c) The coefficient of $x^(11)$ is 120.

Step-by-step explanation:

Remember that the binomial theorem say that $(x+y)^n=\sum_(k=0)^(n) \binom{n}{k}x^(n-k)y^(k)$

a) $(3x+y)^5=\sum_(k=0)^5\binom{5}{k}3^(n-k)x^(n-k)y^k$

Expanding we have that

$\binom{5}{0}3^5x^5+\binom{5}{1}3^4x^4y+\binom{5}{2}3^3x^3y^2+\binom{5}{3}3^2x^2y^3+\binom{5}{4}3xy^4+\binom{5}{5}y^5$

symplifying,

$(3x+y)^5=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5$ .

b) The middle term in the expansion of $((1)/(x) +√(x))^4=\sum_(k=0)^(4)\binom{4}{k}(1)/(x^(4-k))x^{(k)/(2)}$ correspond to k=2. Then $\binom{4}{2}(1)/(x^2)x^{(2)/(2)}=(6)/(x)$ .

c) $(x^2+(1)/(x))^(10)=\sum_(k=0)^(10)\binom{10}{k}x^(2(10-k))(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-2k)(1)/(x^k)=\sum_(k=0)^(10)\binom{10}{k}x^(20-3k)$

Since we need that 11=20-3k, then k=3.

Then the coefficient of $x^(11)$ is $\binom{10}{3}=120$

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