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xxMikexx [17]
3 years ago
10

Suppose we have two thermometers. One thermometer is very precise but is delicate and heavy (X). We have another thermometer tha

t is much cheaper and lighter, but of unknown precision (Y). We would like to know if we can (reliably) bring the lighter thermometer with us into the field. So, we set up an experiment where we expose both thermometers to 31 different temperatures and measure the temperature with each. We get the following observationsx = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120y = 0.02, 3.99, 7.91, 12.03, 16.09, 20.00, 23.98, 28.09, 31.94, 36.03, 40.00, 44.05, 47.95, 52.00, 55.87, 59.90, 63.91, 67.95, 72.11, 76.02, 80.01, 84.10, 88.06, 91.74, 96.02, 99.95, 103.87, 108.01, 111.99, 116.04, 120.03We want to test if these thermometers seem to be measuring the same temperatures. Let's use the threshold . Answer all questions up to 3 decimals(a) Write down the appropriate hypothesis tests for . H0: ---Select--- ≠ = > < and Ha: ---Select--- ≤ = > ≠ ≥ <(b) The test statistic is (Use 2 decimal places)(c) The p-value is (Use 4 decimal places)(d) Therefore, we can conclude thatThe data provides evidence at the 0.1 significance level that these thermometers are not consistentThe data provides no evidence at the 0.1 significance level that these thermometers are not consistentThe probability that the null hypothesis is true is equal to the p-valueThe probability that we have made a mistake is equal to 0.1
Mathematics
1 answer:
Rudiy273 years ago
3 0
Yes you are correct because it really is precise and delicate
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F (x) = 5x + 3<br> What is the value of x when f(x) = 6?
lapo4ka [179]

Answer:

The value of x when f(x) equals 6 is 3/5.

Step-by-step explanation:

In order to solve this problem, we shall start by inputting what we know.

Since the problem provides you with the value of f(x), we will input the value in the given equation.

Original Equation: f(x) = 5x + 3

New Equation: 6 = 5x + 3

Now that all known values of variables have been added to the equation, we will begin to solve.

Start by subtracting both sides of the equation by 3. This step is necessary to isolate x in order to find it's value.

6 = 5x + 3

6 - 3 = 5x + 3 - 3

3 = 5x

Next, we shall divide both side of the equation by 5. This step will allow us to isolate x and finally solve its value.

3 = 5x

3/5 = 5x/5

3/5 = x

Thus, the value of x in f(x) = 5x + 3 is 3/5.

---

To be sure your answer is correct, insert the values of both f(x) and x into the equation provided and solve like so...

f(x) = 5x + 3

6 = 5(3/5) + 3

6 = 3 + 3

6 = 6 ✅

6 0
3 years ago
One golfer's scores for the season are 88, 90, 86, 89, 96, and 85. Another
Allisa [31]

Mean of the first golfer = 89

Mean of the second golfer = 87

Range of the first golfer = 11

Range of the second golfer = 8

Explanation:

First golfer scores 88, 90, 86, 89, 96 and 85.

Sum of the scores of first golfer = 88 + 90 + 86 + 89 + 96 + 85 = 534

Number of observation of first golfer = 6

\text {Mean} = \frac{\text {Sum of the observation}}{\text {Number of observation}}

Mean of the first golfer = \frac{534}{6}=89

Mean of the first golfer = 89

Range of the first golfer = Highest score – Lowest score

                                        = 96 – 85

Range of the first golfer = 11

Second golfer scores 91, 86, 88, 84, 90 and 83.

Sum of the scores of second golfer = 91 + 86 + 88 + 84 + 90 + 83 = 522

Number of observation of second golfer = 6

\text {Mean} = \frac{\text {Sum of the observation}}{\text {Number of observation}}

Mean of the second golfer = \frac{522}{6}=87

Mean of the second golfer = 87

Range of the second golfer = Highest score – Lowest score

                                              = 91 – 83

Range of the second golfer = 8

Comparing the golfer's skills:

Mean of first golfer is greater than Mean of second golfer.  (i.e. 89 > 87)

Range of first golfer is greater than Range of second golfer. (i.e. 11 > 8)

Thus, first golfer have more skills than second golfer.

6 0
3 years ago
Please help. Will reward brainliest
ipn [44]

Answer:

c is the answer


4 0
3 years ago
What is the Volume! ~
Mrrafil [7]

Answer:

<u>108</u>

Step-by-step explanation:

Volume of a Square Pyramid :

  • V = 1/3 x Base Area x Height

Solving :

  • V = 1/3 x (6)² x 9
  • V = 36 x 3
  • V = <u>108</u> cubic centimeters
6 0
2 years ago
Read 2 more answers
The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

    So  

         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

       \r p  =  \frac{210}{500}

       \r p  = 0.42

Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

The critical value of \frac{\alpha }{2} from the normal distribution table is  

      Z_{\frac{\alpha }{2} }  =  2.58

  Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} }*  \sqrt{ \frac{\r p (1- \r p )}{n} }

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

     0.4107 <  p  < 0.4293

 

4 0
3 years ago
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