Statistical data can be classiﬁed into

A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Ba

BartSMP [9]

Answer:

$12.6 < \mu < 14.8$

For this case we can find the margin of error like this:

$ME= \frac{14.8-12.6}{2}= 1.1$

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

$12.6 +1.1=13.7$

$14.8-1.1=13.7$

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=85-1=84$

We know that the confidence interval is given by:

$12.6 < \mu < 14.8$

For this case we can find the margin of error like this:

$ME= \frac{14.8-12.6}{2}= 1.1$

Since we need to divide the width of the interval by 2.

And now with the margin of error we can find the sample mean with any of the two following ways:

$12.6 +1.1=13.7$

$14.8-1.1=13.7$

So for this case the correct answer would be:

A. Sample Mean = 13.7; Margin of Error = 1.1

3 0

3 years ago

What is 71/5 written as a percent

ratelena [41]

Percent (%) = per 100

7 1/5 = 36/5 = 720/100 = 720%

3 0

3 years ago

Name a fourth point in plane TUY. x Y W Z

Elan Coil [88]

Answer:

X

Step-by-step explanation:

6 0

4 years ago

Laura is a single taxpayer, she has $35,000 in ordinary taxable income and $5,000 in capital gains on an investment she held for

Kay [80]

Going by the information given above, Laura is most likely going to belong to the tax bracket where the rate payable is 12%. This is because she is a single filer and her income is between $10,276 and $41,775.

<h3>What is a tax bracket?</h3>

Tax brackets are a series of income and tax layers by which people are defined based on how much they earn per annum.

The tax rates range from 10% to 37%. The rate applicable to a person or group of persons depends on

whether or not they are single
If married, whether they are filing jointly; and
how much their annual income is subject to deductibles.

It is to be noted that a general answer was provided here due to insufficient details in the question.

Learn more about tax brackets at:

brainly.com/question/1615513

#SPJ1

6 0

2 years ago

The nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium. To determine if the

lora16 [44]

Answer:

We conclude that the sodium content is same as what the nutrition label states.

Step-by-step explanation:

We are given that the nutrition label for Oriental Spice Sauce states that one package of sauce has 1100 milligrams of sodium.

The FDA randomly selects 40 packages of Oriental Spice Sauce and determines the sodium content. The sample has an average of 1088.64 milligrams of sodium per package with a sample standard deviation of 234.12 milligrams.

Let $\mu$ = average sodium content.

So, Null Hypothesis, $H_0$ : $\mu$ = 1100 milligrams {means that the sodium content is same as what the nutrition label states}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 1100 milligrams {means that the sodium content is different from what the nutrition label states}

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

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

where, $\bar X$ = sample average sodium content = 1088.64 milligrams

s = sample standard deviation = 234.12 milligrams

n = sample of packages of Oriental Spice Sauce = 40

So, test statistics = $\frac{1088.64-1100}{\frac{234.12}{\sqrt{40}}}$ ~ $t_3_9$

= -0.307

The value of z test statistics is -0.307.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical values of -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test.

Since our test statistics lies within the range of 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 sodium content is same as what the nutrition label states.

3 0

4 years ago