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Yuri [45]
2 years ago
10

PLS PLS ASAP HELP!! 100 POINTS

Mathematics
1 answer:
QveST [7]2 years ago
8 0

The desired statistical measures are given as follows:

1. The margin of error is of M = \frac{6.75}{\sqrt{n}}.

2. The margin of error is of 0.02 = 2%.

3. The point estimate is of 0.655.

<h3>What is the margin of error when we are given the sample standard deviation?</h3>

We also have to be given the sample size n, along with the standard deviation s, and the margin of error is given by:

M = \frac{s}{\sqrt{n}}

In item 1, we have that s = 6.75, hence the margin of error is:

M = \frac{6.75}{\sqrt{n}}

<h3>What is the margin of error when we are given two bounds of an interval?</h3>

The margin of error is given by half the subtraction of the bounds, hence, for item 2:

(0.37 - 0.33)/2 = 0.02.

The margin of error is of 0.02 = 2%.

<h3>What is the point estimate of an interval when we are given two bounds of an interval?</h3>

The points estimate is given by the mean of the bounds, hence, for item 3:

M = (0.61 + 0.70)/2 = 0.655.

The point estimate is of 0.655.

More can be learned about statistical measures at brainly.com/question/25890103

#SPJ1

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Answer:

We conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

Step-by-step explanation:

We are given that a quality-control manager for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates.

Using a random sample of 10 cans, the manager determined that the average calories per can is 124 with a standard deviation of 6 calories.

<u><em>Let </em></u>\mu<u><em> = average calorie content of a 12-ounce can.</em></u>

So, Null Hypothesis, H_0 : \mu \leq 120 calories     {means that the average calorie content of a 12-ounce can is less than or equal to 120 calories}

Alternate Hypothesis, H_A : \mu > 120 calories     {means that the average calorie content of a 12-ounce can is greater than 120 calories}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                         T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average calories per can = 124 calories

             s = sample standard deviation = 6 calories

            n = sample of cans = 10

So, <em><u>test statistics</u></em>  =  \frac{124-120}{\frac{6}{\sqrt{10} } }  ~ t_9

                               =  2.108

The value of t test statistics is 2.108.

<em>Now, at 0.05 significance level </em><em>the t table gives critical value of 1.833 at 9 degree of freedom for right-tailed test</em><em>. Since our test statistics is more than the critical values of t as 2.108 > 1.833, so we have sufficient evidence to reject our null hypothesis as it will in the rejection region due to which </em><u><em>we reject our null hypothesis</em></u><em>.</em>

Therefore, we conclude that the average calorie content of a 12-ounce can is greater than 120 calories.

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