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fredd [130]
3 years ago
9

A public health researcher wishes to study the dietary behavior of residents in Durham County. The researcher randomly contacts

35 county residents and collects data on their daily sugar intake and obtained a sample average of 37.4 grams sugar per day and a sample standard deviation of 4.2 grams per day. The researcher would like to construct a 99% confidence interval for the mean daily sugar intake of residents in the county using the data. Which distribution should the researcher use when analyzing these data?a. The researcher should use a normal distribution because the data come from a simple random sample and the sample is large.b. The researcher should use a normal distribution because the true population average daily sugar intake is likely not near zero.c. The researcher should use a normal distribution because they do not know the true standard deviation of daily sugar intake in Durham county.d. The researcher should use a t-distribution because the sample size is too small to accurately calculate a 99% confidence interval using a normal distribution.e. The researcher should use a t-distribution because they do not know the true standard deviation of daily sugar intake in Durham county.f. The researcher cannot determine whether a t-distribution or a normal distribution would be more appropriate without looking at the shape of the population distribution.
Mathematics
2 answers:
sukhopar [10]3 years ago
5 0

Answer:

(e) The researcher should use a t-distribution because they do not know the true standard deviation of daily sugar intake in Durham county.

Step-by-step explanation:

A t-distribution is used when the true standard deviation (population standard deviation) is unknown. t-distribution uses the standard deviation of the samples.

In this case, it is the sample standard deviation that is known, therefore, the researcher should use a t-distribution.

Sedaia [141]3 years ago
3 0

Answer:

The correct option is option a

The researcher should use a normal distribution because the data come from a simple random sample and the sample is large.

Step-by-step explanation:

The research should use a normal distribution because a simple random distribution was used to get the sample and the sample is large(> = 30).

With the parameters given standard deviation, the sample size, and the population mean, can be used to find the standard score. Which signifies that a normal distribution is the best approach to use.

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Use elimination to solve each system if equation 3x+2y=0and x-5y=17
kotykmax [81]

Answer:

y = -3

x=2

Step-by-step explanation:

3x + 2y = 0

x - 5y=17

to use the elimination you have to make one of the numbers same

so

3(x-5y=17)

and you would have to change the sign so that both of the 3x could cancel out.

New system of equation

3x + 2y = 0

-3x+15y=-51

17y=-51

y = -3

so we know the y value so now you could find the x value take one equation

x - 5y=17

x +15=17

x = 2

you could try it with both of the equations

3x + 2y = 0

3x + - 6=0

3x-6=0

3x=6

x=6/3

x=2

6 0
3 years ago
The branch manager of an outlet (Store 1) of a nationwide
monitta

Answer:

1. ( 19.1416, 23.5384,)

2. (0.276348, 0.46651)

3. the sample size = 170.73 approximately 171

4. sample size = 334.07 approximately 334

5. sample of 334 should be taken by manager

Step-by-step explanation:

mean = bar x = 21.34 dollars

size of sample n = 70

standard deviation of sample = 9.22

we use t distribution as the population standard deviation is not known.

95% Confidence interval

1-α = 0.95

α = 0.05

degree of freedom = 70-1 = 69

α/2 = 0.025

using the t distribution tsble,

= 1.9949

confidence interval = 21.34+-1.9949*[\frac{9.22}{sqrt(70)}] \\

= 21.34 +- (1.9949*1.10200)

= 21.34 + 2.1984, 21.34 - 2.1984

= (23.5384, 19.1416)

the confidence interval of the mean amount spent at the supply store can be written as 19.1416<u<23.5384

2. sixe of those who only have a cat

p = 26/70 = 0.371429

at 90 % confidence interval,

1-α = 0.90

α = 0.10

we use the z table here

z(0.10/2) = Z(0.05)

= 1.645

0.371429+-1.645\sqrt} \frac{0.371429(1-0.371429)}{70}

= 0.371429 +-( 1.645 x 0.0578)

= 0.371429 + 0.095081, 0.371429 - 0.095081

= (0.276348, 0.46651)

3. sd = 10$

margin of erro,r e = 1.50$

α = 0.05

using z table

α/2 = Z0.025

= 1.96

sample size = 1.96² * 10² / 1.50²

= 3.8416 * 100/ 2.25

= 170.73

the sample size is approximately 171

d. we have 0.5 as sample proportion now

margin of error = 0.045

α = 0.10

Zα/2 = 0.05

= 1.645

sample size = 1.645²x0.5(1-0.5) / 0.045²

= 0.676506/0.002025

= 334. 07

sample size = 334

5.  sample of 334 should be taken by manager

3 0
3 years ago
-7 over 1 3rd written as a fraction is
m_a_m_a [10]

Answer: 22/3 I think

sorry if it’s wrong

7 0
2 years ago
Read 2 more answers
Ethan's taxable income is $87,525. Use this tax schedule to calculate the total amount he owes in taxes.
Reptile [31]
$18,617.75 is the answer
$87,525-$77,100=$10,425 (the amount over $77,100)
$10,425 x .28=$2,919 (the 28% taxed amount over $77,100)
$2,919+$15,698.75 (the already taxed amount in the chart)=$18,617.75



8 0
3 years ago
A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co
Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let \mu = <u><em>mean weight of coffee in its containers.</em></u>

SO, Null Hypothesis, H_0 : \mu \geq 32 ounces     {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, H_A : \mu < 32 ounces      {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here <u>One-sample t-test statistics</u> 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 mean weight = 31.8 ounces

             s = sample standard deviation = 0.48 ounces

             n = sample of containers = 15

So, <u><em>the test statistics</em></u>  =  \frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }  ~ t_1_4  

                                      =  -1.614

The value of t test statistics is -1.614.

<u>Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0
3 years ago
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